Please answer this question for points.

16 x 7 x 15 + 11 + 17

Try Ur best and take Ur time.

Answers

Answer 1
The answer for the problem is 18497
Answer 2

Answer:

1,708?

Step-by-step explanation:

pretty sure this is the answer


Related Questions

Identify the surface area of the cylinder to the nearest tenth. Use 3.14 for π.

Answers

Answer:

967.6

Step-by-step explanation:

967.6

967.12 in

Step-by-step explanation:

the formula for the area (surface area) of a cylinder is: A=2πrh+2πr2

to solve we need to determine the values

R= radius = half the diameter = 14/2 =7

D= diameter =14inches

H= height= 15 inches

plug in

A=2πrh+2πr2 = A=2π([tex]\frac{d}{2}[/tex])h+2π([tex]\frac{d}{2}[/tex])^2

= 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])(15) + 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])^2

= 2[tex]\pi[/tex](7)(15) + 2[tex]\pi[/tex](7)^2

=[tex]\pi[/tex]((2x7x15)+(2x7^2))

=[tex]\pi[/tex](210+98)

=[tex]308\pi[/tex]

=967.12 in

HURRY PLEASE! I NEED HELP. PLEASE GIVE ME THE ANSWER

Use the table to describe the function.

A 2-column table with 6 rows. Column 1 is labeled x with entries negative 1,000, negative 0.1, negative 0.0001, 0.0001, 0.1, 1,000. Column 2 is labeled f (x) with entries negative 1.9999982, 1.78, 1.799998 times 10 Superscript 6 Baseline, 1.799998 times 10 Superscript 6 Baseline, 1.78, negative 1.9999982.

The end behavior of the function is that as
x →±∞, y approaches
.

The Limit of f (x) as x approaches infinity
.

The function has an asymptote at
.

Answers

Answer:

-2

-2

y=-2

Step-by-step explanation:

Answer:

-2

-2

y=-2

Step-by-step explanation:

The scale factor for a model is 8 cm m Model : 11.23cm actual: 40.2 m​

Answers

the answer is = 19 m

Which value is not a solution of the inequality x-4 symbol 2

Answers

The inequality x -4 > 2 uses a greater than symbol

All numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

How to determine the value not in the solution?

The inequality is given as:

x -4 > 2

Add 4 to both sides of the inequality

x - 4 + 4 > 2 + 4

Evaluate the sum

x > 6

The above means that only numbers greater than 6 are in the solution of the inequality.

Since the options are not given, I will give a general solution that all numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

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∠A and \angle B∠B are vertical angles. If m\angle A=(7x-6)^{\circ}∠A=(7x−6)



and m\angle B=(8x-27)^{\circ}∠B=(8x−27)



, then find the measure of \angle B∠B

Answers

keeping in mind that vertical angles are always congruent.

[tex]\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}[/tex]

Find the circumference of the circle of 13 inches use 3.14 for pi and round to the nearest whole number

Answers

Answer:

ohh just use this formula

this is for Area- A= π r^2

this is for Circumference- C= 2 π r

Step-by-step explanation:

The circumference of the circle of the radius of 13 inches is, 41 inches

What is circumference ?

Circumference is the distance around the perimeter of a circular object. It is defined as the length of the circle that is found by multiplying the diameter of the circle by π (pi), which is approximately equal to 3.14.

The formula for the circumference of a circle is given by: C = 2πr,

Given that,

The diameter of the circle is 13 inches,

The radius of the circle can be calculated as follows:

r = d/2

  = 13 inches / 2

  = 6.5 inches

Using the formula for the circumference, we can calculate the circumference as follows:

C = 2πr

   = 2 × 3.14 × 6.5 inches

   = 40.76 inches

Rounding to the nearest whole number, the circumference of the circle is 41 inches.

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help me please need help​

Answers

Answer:

Step-by-step explanation:

1.  x -> opposite side of 48°

   o → hypotenuse

   b → adjacent side of  48°

[tex]\sf Sin \ 48^\circ = \dfrac{opposite \ side }{hypotenuse}\\\\\\0.7431 = \dfrac{15}{o}\\\\\\0.74 * o = 15\\\\\\ o = \dfrac{15}{0.74}\\\\\\[/tex]

o = 20.27

[tex]\sf cos \ 48^\circ = \dfrac{adjacent \ side }{hypotenuse}\\\\\\0.67 =\dfrac{b}{o}\\\\\\0.67=\dfrac{b}{20.27}[/tex]

b = 0.67*20.27

b = 13.58

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2) i → opposite side of 25°

n → adjacent side of 25°

[tex]\sf Sin \ 25 =\dfrac{i}{t}\\\\\\0.42=\dfrac{i}{30}\\\\\\0.42*30=i[/tex]

i = 12.6

[tex]\sf Cos \ 30^\circ =\dfrac{n}{t}\\\\0.91=\dfrac{n}{30}\\\\\\0.91*30 = n[/tex]

n = 27.3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3) a → opposite side of 70°

e → adjacent side of 70°

[tex]Sin \ 70^\circ =\dfrac{a}{l}\\\\0.94 =\dfrac{a}{25}\\\\0.94*25=a[/tex]

a = 23.5

[tex]\sf Cos \ 70^\circ =\dfrac{e}{l}\\\\0.34=\dfrac{e}{25}\\\\0.34*25=e[/tex]

e = 8.5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4)

[tex]\sf Sin \ 52^\circ = \dfrac{x}{75}\\\\0.79*75=x\\[/tex]

x = 59.25

[tex]\sf Cos \ 52^\circ = \dfrac{z}{75}\\\\0.62*75 =z[/tex]

z = 46.5

What are the solutions of the equation (x + 2)2 + 12(x + 2) – 14 = 0? Use u substitution and the quadratic formula to
solve.

-8+5√2
O x=-6252
O x=-4+5√2
x=-2 +5√2

Answers

Answer:

-8+5√2

Step-by-step explanation:

(x+2)^2+12(x+2)–14=0

(x+2)^2=(x+2)(x+2)=x^2+4+4x

12(x+2)=12x+24

x^2+4+4x+12x+24-14=0

x^2+4x+12x+4+24-14=0

x^2+16x+14=0

quadratic formula

x = {-b +- square root of (b^2 – 4ac)} ÷ {2a}

a= 1

b = 16

c = 14

x = {-16 +- square root of (16^2 – 4*1*14)} ÷ {2*1}

x = {-16 +- square root of (256 – 56)} ÷ {2*1}

x = ((-16 +- square root of (200)) ÷ (2)

x = ((-16 +- 10√2)) ÷ (2)

x= -8+-5√2

Find the values that will make the equations true.

options:

a = 49


a = 27.5


a = 7


a = 24.5

Answers

The value of a=7 will make the equation true

Which pair of expressions has equivalent values?
1^13 and 1^15
6^1and 9^1
7^8and 8^7
9- and 4^3

Answers

Answer:

1^13 and 1^15

Step-by-step explanation:

1 raised to anything is still just 1

so, 1^13 = 1 and 1^15 =1

Tom wants the scale model to be 9 inches tall.
How wide should the scale model be?
A. 1.7 inches
B. 5.4 inches
O c. 15 inches
OD. 20 inches

Answers

Answer:

C 15

Step-by-step explanation:

9 =  12 3/4

20 3/4 = 15

What are range, index of qualitative variation (IQV), interquartile range (IQR), standard deviation, and variance

Answers

Answer:

To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

enter the value of c when the expression 21.2x+c is equivalent to 5.3(4x-2.6). PLEASE GIVE ME AN EXPLANATION AS WELL. I DON’T WANT JUST AN ANSWER.

Answers

Well, I will not just give you a straight answer, since I do not have the resources to currently do so. But as a tip, I would try the substitution method to get rid of the ‘x’ variable and just solve for ‘c’ like normal. Your main expression should look like this, I believe:

21.2x+c = 5.3(4x-2.6)

To substitute h the is out, you could write this if it feels better for you:

21x+c = y and 5.3(4x-2.6) = y

Hope this helps!

Which expression is equal to 0.75×0.09

Answers

The answer is C 75/100 * 9/10

Samir recorded the grade-level and instrument of everyone in the middle school School of Rock below. Seventh Grade Students Instrument # of Students Guitar 6 Bass 4 Drums 6 Keyboard 7 Eighth Grade Students Instrument # of Students Guitar 9 Bass 9 Drums 9 Keyboard 10 Based on these results, express the probability that a seventh grader chosen at random will play an instrument other than drums as a fraction in simplest form.

Answers

Using it's concept, it is found that there is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

There is a total of 6 + 4 + 6 + 7 = 23 seventh graders.Of those, 23 - 6 = 17 play instruments that are not the drum.

Hence:

[tex]p = \frac{17}{23}[/tex]

There is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

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8. Daisies and tulips are planted in a
garden. There are 11 fewer tulips
planted than daisies.
a. Write an expression that represents
the number of tulips in terms of the
number of daisies. Define any
variables used.
b. If 18 daisies are planted, how many
tulips are planted?

Answers

Answer:

a) [tex]d-11 = t[/tex]

b) 7

Step-by-step explanation:

Let [tex]d[/tex] = daisies

Let [tex]t[/tex] = tulips

a) 11 fewer tulips than daisies: [tex]d-11 = t[/tex]

b) Substitute 18 into [tex]d[/tex] and solve for [tex]t[/tex].

[tex]18-11= t[/tex]

[tex]7 = t[/tex]

Mr. Fuller wants to put fencing around his rectangular-shaped yard. the width of the yard is 55 feet and the length is 75 feet. how many feet of fencing does Mr. Fuller

Answers

If mr fuller wants to put up fencing you need to find the area

The total length of the fencing is the perimeter of the rectangular yard which is P = 260 feet

What is the Perimeter of a Rectangle?

The perimeter P of a rectangle is given by the formula, P=2 ( L + W) , where L is the length and W is the width of the rectangle.

Perimeter P = 2 ( Length + Width )

Given data ,

Let the perimeter of the rectangular yard be = P

Now , the equation will be

The width of the rectangular yard W = 55 feet

The length of the rectangular yard L = 75 feet

And , the required fencing = Perimeter of rectangular yard

Perimeter of rectangular yard P = 2 ( L + W )

Substituting the values in the equation , we get

Perimeter of rectangular yard P = 2 ( 55 + 75 )

Perimeter of rectangular yard P = 2 ( 130 )

Perimeter of rectangular yard P = 260 feet

Therefore , the value of P is 260 feet

Hence , the perimeter of yard is 260 feet

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Prove that: a + b + c / a^-1+ b^-1+ c^-1 = abc


Answers

Answer:

It's right

Step-by-step explanation:

(dk how to show prove but thank?

7. What value of c will make x2 – 20x + c
a perfect square trinomial?

Answers

29x for this answer

I need this for school, please help!!

Answers

C. Firstly, add up the amount of students and you should get 352. I did the easy route and divided 352 by 8 and got 44. From there, I added the amount of teachers on each graph and C is the only one with the amount of 44 teachers.

Which expression shows the factored form of 15x + 4

Answers

It doesn’t factor. If you draw it on graph its just a straight line with only 1 x intercept

Answer:

Step-by-step explanation:

The only factor you can use is H. This doesn't factor into anything nice (like the common factor would be a decimal which is usually not allowed when doing this kind of question.

Answer: The question is the answer. 15x + 4 or H

A kitchen can be broken into 2 rectangles. One rectangle has a base of 7 feet and height of 5 feet. The second rectangle has a base of 2 feet and height of 2 feet. One package of tile will cover 3 square feet. How many packages of tile will she need? 8 13 15 39

Answers

Answer:

its 13 or B

Step-by-step explanation:

Number 21 please help me solve it thank youu

Answers

Answer:

64 pi

Step-by-step explanation:

32 is diameter

diameter is circumference

2 circles

so 2*32=64

NEED HELP ASAP!!!! Will give brainiest

Answers

2%(random sentence bc I need 20 characters)

ASAP PLS
Write an equation to represent the following scenario: Ms. Cloutier’s wedding photographer requires a $1000 deposit, and then $250 for every hour she is working.

Answers

1000+(250•x)= y

x being the number of hours she’s working

Hello Calculus!

Find the value

[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_3^5}}(e^{3x}+7cosx-3tan^3x)dx[/tex]

Note:-

Answer with proper explanation required and all steps to be mentioned .

Answers

Answer:

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

General Formulas and Concepts:
Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Rule - Addition/Subtraction]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + \int\limits^5_3 {7 \cos x} \, dx - \int\limits^5_3 {3 \tan^3 x} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^3 x} \, dx[/tex][3rd Integral] Rewrite:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution and u-solve.

1st Integral

Set u:
[tex]\displaystyle u = 3x[/tex][u] Differentiate [Derivative Properties and Rules]:
[tex]\displaystyle du = 3 \, dx[/tex][Bounds] Swap:
[tex]\displaystyle \left \{ {{x = 5 \rightarrow u = 3(5) = 15} \atop {x = 3 \rightarrow u = 3(3) = 9}} \right.[/tex]

3rd Integral

Set v:
[tex]\displaystyle v = \sec x[/tex][v] Differentiate [Trigonometric Differentiation]:
[tex]\displaystyle dv = \sec x \tan x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, dv[/tex]

Step 4: Integrate Pt. 3

Let's focus on the 3rd integral first.

Apply Integration Method [U-Solve]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \int\limits^{x = 5}_{x = 3} {\frac{v^2 - 1}{v}} \, dv[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \int\limits^{x = 5}_{x = 3} {v} \, dv - \int\limits^{x = 5}_{x = 3} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Integration Rules [Reverse Power Rule and Logarithmic Integration]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 5}_{x = 3} - \ln | v | \bigg| \limits^{x = 5}_{x = 3} \Bigg)[/tex][v] Back-Substitute:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 x}{2} \bigg| \limits^{5}_{3} - \ln | \sec x | \bigg| \limits^{5}_{3} \Bigg)[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 5 - \sec^2 3}{2}- \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \Bigg)[/tex]

Step 5: Integrate Pt. 4

Focus on the other 2 integrals and solve using integration techniques listed above.

1st Integral:

[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]

2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]

Step 6: Integrate Pt. 5

[Integrals] Substitute in integrals:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

∴ we have evaluated the integral.

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---

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer:

1086950.36760

Formula's used:

[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]

[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]

[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]

[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]

[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]

Explanation:

[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]

                        apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]

[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]

                Integrate simple followings first, using formula's given above

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]

                        Breakdown the component

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]

                                                         [ tan²x = sec²x - 1 ]

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]

===========================================================

for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]

                                                  apply substitution ... u

[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]

[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]

[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]

[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]

substitute back u = sec(x)

[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]

================================================= insert back

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex]   outcome after integrating

Now apply the given limits

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                                                                   simplify

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                          and group the variables

[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]

value:

[tex]\sf \hookrightarrow 1086950.36760[/tex]

When one variable precedes the other in time and the correlation between them is high, which statement can be made with confidence?
A) The variable that comes earliest in time causes the other variable
B) There is a high level of association between the two variables
C)The two variables have an association with an outcome variable
D)There are no extraneous variables at work in the situation

Answers

Using correlation coefficients, it is found that the correct option is given by:

B) There is a high level of association between the two variables.

What is a correlation coefficient?

It is an index that measures correlation between two variables, assuming values between -1 and 1.If it is positive, the relation is positive, that is, they are direct proportional. If it is negative, they are inverse proportional.If the absolute value of the correlation coefficient is greater than 0.6, the relationship is strong.

In this problem, it is said that the correlation between them is high, that is, strong, hence there is a high level of association, which means that option B is correct.

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What is 40 x 40 x 40 please help fast ASAP

Answers

40^3 or 4 times 4 times 4 times 10 times 10 times 10
Which is
64 times 1000
Which is
64000

A right triangle includes one algae that measures 14º. what is the measure of the third angle

A 14º C 90º

B 76º D 104º

Answers

Answer: B. 76

Step-by-step explanation: A right triangle is 180 degrees. It has an angle of 90 since it is a right triangle. 90 + 14 = 104. 180 - 104 = 76

Answer:

B. 76 degrees

Step-by-step explanation:

EVERY triangle's angles add up to 180 degrees. We already know that since it's a right triangle, one of the angles equals 90 degrees (that's a right angle) and they give us the second angle measurement, 14 degrees. If we add those two angle measures together and subtract them from 180, we should get the measure of the third angle as our answer.

14 + 90 = 104

180 - 104 = 76

Therefore, the third and final angle in this right triangle equals 76, so your answer is B. I hope this helps! Have a lovely day!! :)

Need help with this problem

Answers

Answer:

A)

Step-by-step explanation:

Sum of all angles of triangle = 180

53 + 68  + x = 180

        121 + x  = 180

                 x = 180 - 121

                 x = 59°

Answer:

180 - 53 - 68 = 59

The third angle is 59°.

Step-by-step explanation:

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