7th grade math and serface area

7th Grade Math And Serface Area

Answers

Answer 1

Answer:

108 square yards

Step-by-step explanation:

l = 6 yd

w = 4 yd

h = 3 yd

Surface  area of rectangular prism = 2*(lw +wh + hl)

                                                            = 2*(6*4 + 4*3 + 3*6)

                                                            = 2* ( 24 + 12 + 18)

                                                            = 2 *54

                                                            = 108 square yards


Related Questions

340 to the nearest 100

Answers

Answer:

300

Step-by-step explanation:

It’s 300 because on the rounding scale, 4 means to round lower. Hope this helps.

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:

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

i doont understand??

Answers

The person above ⬆️ is correct
Give him brainliest <3

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

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

Answers

29x for this answer

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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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

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

A football field is 120 yards long by 53yr wide. If a player runs diagonally from one corner to the opposite corner, how far will they travel

Answers

Answer:

They will travel about 131.18 yards.

Step-by-step explanation:


The ratio 125 : x is equivalent to the ratio x^2 : 125. What is the value of x?

Answers

Answer:

  x = 25

Step-by-step explanation:

The proportion can be written using fractions and solved for x in the usual way.

  [tex]\dfrac{125}{x}=\dfrac{x^2}{125}\\\\125^2=x^3\qquad\text{multiply by 125x}\\\\x=\sqrt[3]{125^2}=\sqrt[3]{5^6}=5^2\qquad\text{take the cube root}\\\\\boxed{x=25}[/tex]

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:

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

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?

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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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

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

Answers

the answer is = 19 m

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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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

Hi, I think the domain is 5 am I right?

Answers

Its 4.

Starts at -1 and goes to 3

3+1 =4

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.

The quantity z varies directly with w and inversely with x. When w = 12 and x = 4, z = -42. Find z when w = 11 and x = 7.


PLEASE HELP ME!!!!!!!

Answers

Given the joint variation and the constant of variation, the value of z when w = 11 and x = 7 is -22

How to solve variation?

z =( k × w) / x

where,

z = -42 w = 12x = 4k = constant of variation

-42 = (k × 12) / 4

cross product

4 × -42 = 12k

-168 = 12k

k = -168 /12

k = -14

Find z when w = 11 and x = 7

z =( k × w) / x

z = (-14 × 11) / 7

= -154 / 7

z = -22

Therefore, the value of z when w = 11 and x = 7 is -22

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The volume of a right cylinder is 108π and its height is 12.

What is the length of the cylinder's radius?

Answers

Answer:

volume of a cylinder = π r^2 *h

108π=π*r^2*h

108=r^2*h

12=r^2

3.464=r

radius =3.464

Step-by-step explanation:

Brainliest would be appreciated

Ask for questions!

Answer:

The length of the cylinder's radius is 3

Step-by-step explanation:

A cylinders radius is found by the formula, r= √ v/πh

So, 108π/πh.

Because there is pi in both the numerator and denominator, we can cancel them out. This leaves the equation to be : r = √ 108/12.

r = √ 108/12

108/12 is 9.

r = √9

r = 3

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:

Which expression is equal to 0.75×0.09

Answers

The answer is C 75/100 * 9/10

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

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:

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!

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

Other Questions
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