F
E
58>D
C
128
B
128

Answer:

Trim A

Inches = 36

Feet = 3

Yards = 1

Trim B

Inches = 216

Feet = 18

Yards = 6

Answer:

$3,640

Step-by-step explanation:

8.2% as a decimal is 0.082. multiply 2000 by 0.082, and you get your interest rate/the interest you pay each year, which would be 162. multiply 162 by ten to get your interest for all ten years, since 162 was simply the rate, and you get 1,640. add the original 2000 onto that, and you get 3,640.

Answer:

x= -5

Step-by-step explanation:

3x+3=2x-2

3x-2x=-3-2

x=-5

Answer:

x = -5

Step-by-step explanation:

First Expand the equation

3x+3 = 2x-2

Subtract 2x from both sides

x+3=-2

subtract 3 from both sides

x = -5

Hope this helps :)

Answer:

V ≈ 5144.9 cubic units

Step-by-step explanation:

SA(sphere) = 4πr²

1156π = 4πr²

π's cancel out

1156 = 4r²

r² = 1156/4

r² = 289

r = √289

r = 17

V(sphere) = 4/3πr³

= 4/3 × π × 17³

= 4/3π × 4913

V ≈ 5144.9 cubic units

Answer:

[tex]h = -42[/tex]

Step-by-step explanation:

First, subtract 54 from both sides.

[tex]h = 4^2 - 16 \div 4 - 54[/tex]

Then, evaluate the right-hand expression using the order of operations:

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

We'll first evaluate the exponent in 4²:

[tex]4^2 = 4 \times 4 = 16[/tex]

[tex]h = 16 - 16 \div 4 - 54[/tex]

Next, we can evaluate the division:

[tex]16 \div 4 = 4[/tex]

[tex]h = 16 - 4 - 54[/tex]

Finally, perform the subtraction.

[tex]16 - 4 - 54 = -42[/tex]

[tex]h = -42[/tex]

Answer:

[tex] \sf \: h = - 42[/tex]

Step-by-step explanation:

Given equation,

→ h + 54 = 4² - 16 ÷ 4

Now the value of h will be,

→ h + 54 = 4² - 16 ÷ 4

→ h + 54 = 16 - 4

→ h + 54 = 12

→ h = 12 - 54

→ [ h = -42 ]

Hence, value of h is -42.

Spanish killed nearly 6000 Incas.

The Spanish conquistador and explorer Francisco Pizarro springs a trap on Atahualpa, the monarch of the Incas, in the year 1532. Atahualpa allowed roughly 6,000 unarmed men to attend the feast despite having almost 80,000 soldiers with him in the mountains. It took the Spanish a long time and much violence to conquer the Inca Empire. Pizarro's soldiers massacred the 6,000 Incans in approximately one hour.

The lone Spanish wounded were received by Pizarro himself, who injured his hand while saving Atahualpa's life. Pizarro retained Atahualpa in captivity while he prepared plans to conquer his empire because he understood that the emperor was worthy alive than dead. Atahualpa responded by appealing to his hostages' avarice and supplying them with money and silver in exchange for their release.

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Answer:

B

Step-by-step explanation:

the sum of the two angles is 180

30 + 15x = 180

15x =150

x = 10 degrees

Answer:

5x−4y

Step-by-step explanation:

:D

Answer: 5x - 4y

Step-by-step explanation: combining like terms means to combine the terms that have the same letter. so since the 3 and 2 have an x you combine them and the 3 and -7 have a y you combine them. hope this helped:)

Answer:

y intercept:  -2

Step-by-step explanation:

hope this helps :)

Answer:

Slope=4 and the intercept on the y-axis is -2

Step-by-step explanation:

Use the formula y=mx+c

m=slope

c is the intercept on the y-axis

y=4x-2 can be like this

Y=4x+(-2)

That is why slope is 4 and intercept on y-axis is -2

:)

Answer:

B

Step-by-step explanation:

Theorem: If a quadrilateral has 2 sets of opposite sides congruent, 2 sets of opposite angles congruent, and has consecutive angles which are supplementary, then it is a parallelogram.

Hope this helps!

According to the given algebra Darla is 17 years old and Maria is 30 years old.

Numbers, parameters, constants, expressions, equations, linear equations, and quadratic equations are all part of the fundamentals of algebra. Additionally, the algebraic expressions contain the fundamental arithmetic operations of addition, subtraction, multiplication, as well as division.

Maria is 4 years younger than Don,

thus if Don is x years old, then Maria is x years younger. (2x-4)

If their ages added together equal 47,

then x + (2x-4) = 47

Solving for x, then determining Maria's age (2x-4)

3x=47+4

3x=51

x=51/3

x=17 years

Maria's age=2*17-4=30 years

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Answer:

d

a

Step-by-step explanation:

f(3)= 3(3)-1

f(3)=9-1=8

f(x+1)= (x+1)^2-3(x+1)+2

=x^2+2x+1-3x-3+2

=x^2-x

Answer:11 weekends

Step-by-step explanation:

it will take 11 weekends

We know that :

⊕   Area of a Rectangle is given by : Length × Width

Let us calculate the Area of the Rectangle in the Left side of the figure

⇒   Area of Left Rectangle : (x - 3) (x + 2x)

⇒   Area of Left Rectangle : (x - 3) (3x)

⇒   Area of Left Rectangle : 3x² - 9x

Let us calculate the Area of the Rectangle in the right side of the figure

⇒   Area of the Right Rectangle : (x - 1)(x)

⇒   Area of the Right Rectangle : x² - x

Given : Combined area of both rectangles is 50 cm²

⇒   3x² - 9x + x² - x = 50

⇒   4x² - 10x = 50

⇒   2x² - 5x - 25 = 0

Comparing the above equation with ax² + bx + c = 0,

we can notice that :

→   a = 2

→   b = -5

→   c = -25

Answer:

a = 191/81

Step-by-step explanation:

a+ 1 1/6 = 11 7/9

1 1/6 = 7/6

11 7/9 = 106/9

So, our equation is

a + 7/6 = 106/9

Subtract 7/6 from both sides

a = 191/81

So, the answer is

a = 191/81

The function be f(x) = (6) then growth factor for the given equation (b) = 6.

Quantity grows exponentially over time. It happens when a quantity's instantaneous rate of change with regard to time is proportional to the quantity itself.

A function called exponential growth illustrates an expansion within a population that happens at the same pace across time. When a population's per capita growth rate remains constant across time, regardless of population size, exponential growth occurs, causing the population to grow exponentially as the population increases.

The general exponential growth function is given by :-

[tex]$f(t)=A b^t$[/tex] , where A is the initial amount , b is the growth factor and t is the time period.

The given function : [tex]$f(x)=(6)^x$[/tex]

When we compare it to the general exponential equation , we get

The growth factor for the given equation (b) = 6.

The complete question is:

Consider the function f(x) = (6)x. What is the value of the growth factor of the function?

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The function be f(x) = (6) then growth factor for the given equation (b) = 6.

Quantity grows exponentially over time. It happens when a quantity's instantaneous rate of change with regard to time is proportional to the quantity itself.

A function called exponential growth illustrates an expansion within a population that happens at the same pace across time. When a population's per capita growth rate remains constant across time, regardless of population size, exponential growth occurs, causing the population to grow exponentially as the population increases.

The general exponential growth function is given by :-

F(t) = AB^t, where A is the initial amount , b is the growth factor and t is the time period.

The given function : f(t) = (6)^x

When we compare it to the general exponential equation , we get

The growth factor for the given equation (b) = 6.

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The complete question is:

Consider the function f(x) = (6)x. What is the value of the growth factor of the function?

The amount he will have at the end of 3 years is $1596.55.

Option B is the correct answer.

It is the interest we earned on the interest.

The formula for the amount earned with compound interest after n years is given as:

A = P [tex](1 + r/n)^{nt}[/tex]

P = principal

R = rate

t = time in years

n = number of times compounded in a year.

We have,

P = $1220

r = 9%

n = 12

t = 3 years

Amount = P [tex](1 + r/n)^{nt}[/tex]

A = 1220 [tex](1 + 9/12)^{36}[/tex]

A = 1220 [tex](1 + 0.75)^{36}[/tex]

A = 1220 x [tex]1.75^{36}[/tex]

A = $1,596.55

Thus,

The amount after 3 years is $1596.55.

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Answer:

rectangular prism maybe

Step-by-step explanation:

The reasons and responses to the graphs in the question are as follows:

Graph 1.

The amplitude of the first graph is (1/2) which is the height from the midline to the peak

The value of the function at x = 0 is -(1/2), where sin(0) = 0, therefore, the function is a cosine function

The period of the graph is 2·π, which is the period of the parent cosine function

Therefore, the correct option is

Graph 2: Please find attached the graph of the function g(x) = 2×cos(x)

Graph 3: The frequency of a sinusoidal function is given as follows;

The period of the graph = π

Therefore;

The frequency of the sinusoidal graph is 2

Graph 4.  Required:

To find the equation that represents the function of the graph

Solution;

The period of the function, T = π

The graph of the function has a maximum at x = 0, therefore, the graph is similar to a cosine function, y = cos(B·x)

Where;

B = 2·π/T

Therefore;

B = 2·π/π = 2

Therefore;

The equation that represent the function in the graph is f(x) = cos(2·x)

Question 5. The given function is f(x) = cos(2·x)

The frequency factor in the given function, B = 2

The period, T = 2·π/B

Therefore, T = 2·π/2 = π

The period of the function, f(x) = cos(2·x), is 2

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Answer:

Everything seems correct, I'm assuming you wanted confirmation

[tex]sin(A)+cos(A)=\sqrt{2}\hspace{10em}tan(A)+cot(A)=2 \\\\[-0.35em] ~\dotfill\\\\ tan(A)+cot(A)=2\implies \cfrac{sin(A)}{cos(A)}+\cfrac{cos(A)}{sin(A)}=2 \\\\\\ \cfrac{sin^2(A)+cos^2(A)}{cos(A)sin(A)}=2\implies \boxed{\cfrac{1}{cos(A)sin(A)}=2} \\\\[-0.35em] ~\dotfill[/tex]

[tex]sin(A)+cos(A)=\sqrt{2}\implies (~~sin(A)+cos(A)~~)^2=(\sqrt{2})^2 \\\\\\ sin^2(A)+2sin(A)cos(A)+cos^2(A)=2 \\\\\\ 2sin(A)cos(A)+sin^2(A)+cos^2(A)=2\implies 2sin(A)+1=2 \\\\\\ 2sin(A)cos(A)=1\implies \boxed{2=\cfrac{1}{cos(A)sin(A)}}[/tex]

now, another way to look at this identity will  be as a unified system of equations

[tex]\begin{cases} (~~sin(A)+cos(A)~~)^2=2\\\\ ~~ ~tan(A)+cot(A)=2 \end{cases}\implies (~~sin(A)+cos(A)~~)^2=tan(A)+cot(A)[/tex]

and we'd end up with the same rigamarole.

Answer:

1. 170.083 in³

2. 126π in³

3. 92.106 m³

4. 2412.74 in³

5. 612π m³ and 1922 m³

Step-by-step explanation:

1.

Cylinder:

[tex]V = \pi r^{2}h[/tex]               *Plug in numbers*

[tex](3.14)(2.5)^{2}(7)[/tex]         *Square 2.5*

[tex](3.14)(6.25)(7)[/tex]        *Solve*

≈ [tex]137.375in^{3}[/tex]

Sphere:

[tex]V = \frac{4}{3}\pi r^{3}[/tex]              *Plug in numbers*

[tex]\frac{4}{3} (3.14)(2.5)^{3}[/tex]         *Cube 2.5*

[tex]\frac{\frac{4}{3}(3.14)(15.625)}{2}[/tex]         *Divide by 2 and Solve*

≈ [tex]32.7083 in^{3}[/tex]

Add both volumes

[tex]137.375 + 32.7083[/tex] ≈ [tex]170.083in^{3}[/tex]

2.

Cylinder:

[tex]V = \pi r^{2}h[/tex]         *Plug in numbers*

[tex]\pi (3)^{2}(10)[/tex]            *Square 3*

[tex]\pi (9)(10)[/tex]             *Multiply*

[tex]90\pi[/tex]

Sphere:

[tex]V = \frac{4}{3}[/tex] π [tex]r^{3}[/tex]            *Plug in numbers*

[tex]\frac{4}{3}\pi (3)^{3}[/tex]                   *Cube 3*

[tex]\frac{4}{3} \pi (27)[/tex]                   *Multiply*

[tex]36\pi[/tex]

Add both Volumes to get total

[tex]90\pi + 36\pi = 126in^{3}[/tex]

3.

Sphere:

[tex]V = \frac{4}{3}\pi r^{3}[/tex]              *Plug in numbers*

[tex]\frac{4}{3} (3.14)(3)^{3}[/tex]            *Cube 3*

[tex]\frac{4}{3} (3.14)(27)[/tex]            *Multiply*

[tex]113.04m^{3}[/tex]

Cone:

[tex]V = \frac{\pi r^{2}h}{3}[/tex]             *Plug in numbers*

[tex]\frac{(3.14)(2)^{2}(5)}{3}[/tex]           *Square 2*

[tex]\frac{(3.14)(4)(5)}{3}[/tex]             *Solve*

[tex]20.93m^{3}[/tex]

Subtract the volumes to get the volume of the blue area

[tex]113.04 - 20.93 = 92.106m^{3}[/tex]

4.

Sphere:

[tex]V = \frac{4}{3} \pi r^{3}[/tex]            *Plug in numbers*

[tex]\frac{4}{3}\pi (8)^{3}[/tex]                 *Cube 8*

[tex]\\\frac{4}{3}\pi (512)[/tex]               *Multiply*

[tex]\\\\\pi (682.6)[/tex]              *Solve*

[tex]2133.66in^{3}[/tex]           *Divide by 2 since it's a hemisphere*

Cone:

[tex]V = \frac{\pi r^{2}h}{3}[/tex]            *Plug in numbers*

[tex]\frac{\pi (8)^{2}(20)}{3}[/tex]              *Square 8*

[tex]\frac{\pi (64)(20)}{3}[/tex]              *Multiply and Divide*

[tex]1340.41 in^{3}[/tex]

Add both volumes

[tex]1072.33 + 1340.41 = 2412.74in^{3}[/tex]

5.

Cylinder:

[tex]V = \pi r^{2}h[/tex]            *Plug in numbers*

[tex]\pi (6)^{2}(16)[/tex]              *Square 6*

[tex]\pi (36)(16)[/tex]             *Multiply*

[tex]576\pi[/tex]

Cone:

[tex]V = \frac{\pi r^{2}h}{3}[/tex]            *Plug in numbers*

[tex]\frac{\pi (6)^{2}3}{3}[/tex]                  *Square 6*

[tex]36\pi[/tex]

Add both volumes

[tex]576\pi + 36\pi = 612\pi m^{3}[/tex]

Alternative: *Multiply π*

[tex]1922m^{3}[/tex]

Answer:

1) 175 [tex]in^2[/tex]

2) 1st option

3) 92 [tex]m^2[/tex]

4) 2412 [tex]in^3\\[/tex]

5) 2nd option

Step-by-step explanation:

[tex]1)\ A = D * Pi = 5 * 3.14 = 15.7\\[/tex]

[tex]2)\ V_{1} = 15.7 * 7 = 109.9[/tex]

[tex]3)\ V_{2} = 4Pi*\frac{R^3}{3} = 12.56*\frac{15.625}{3} = 65.41(6)[/tex]

[tex]4)\ V_1 + V_2 = 65.4 + 109.9 = 175.3[/tex] which is close to 175

[tex]1)\ A = Pi * R^2 = 3.14 * (\frac{6}{2})^2 = 3.14 * 3^2 = 28.26\\[/tex]

[tex]2)\ V_1 = A * h = 28.26 * 10 = 282.6[/tex]

[tex]3)\ V_2 = \frac{4}{3} * Pi * R^3 = \frac{4}{3} * 3.14 * (\frac{6}{2})^3 = 113.04[/tex]

[tex]4)\ V_1 + V_2 = 282.6 + 113 = 395.6[/tex] which is very close to 396

[tex]5)\ 396 = 126 * pi[/tex]

[tex]1) V_1 = \frac{1}{3} * Pi * R^2 * h = \frac{1}{3} * 3.14 * 2^2 * 5 = 20.9(3)[/tex] =

[tex]2)\ V_2 = \frac{4}{3} * Pi * R^3 = \frac{4}{3} * 3.14 * 3^3 = 113.04\\[/tex]

[tex]3)\ V_2 - V_1 = 113.04 - 20.93 = 92.14[/tex] which is very close to 92

[tex]1)\ V_1 = \frac{\frac{4}{3} * Pi * R^3}{2} = \frac{\frac{4}{3} * 3.14 * 512}{2} = 1,071.786\\[/tex]

[tex]2)\ V_2 = \frac{1}{3} * Pi * R^2 * h = \frac{1}{3} * 3.14 * 64 * 20 = 1,339.733[/tex]

[tex]3)\ V_1 + V_2 = 1072 + 1340 = 2,411 = 2412[/tex]

[tex]1)\ A = Pi * R^2 = 36*Pi\\[/tex]

[tex]2)\ V_1 = 36*Pi * 16 = 576 * Pi\\[/tex]

[tex]3)\ V_2 = \frac{1}{3} * Pi * R^2 * h = \frac{1}{3} * Pi * 36 * 3 = 36 * Pi[/tex]

[tex]4) V_1 + V_2 = 576Pi + 36Pi = 612Pi[/tex]

An equation that could be used to determine the number of wreaths that Nicole could decorate using one spool of ribbon is:

                           W = S / R

The definition of an equation in algebra is a mathematical statement that demonstrates the equality of two mathematical expressions. For instance, the equation 3x + 5 = 14 consists of the two equations 3x + 5 and 14, which are separated by the 'equal' sign.

An algebraic expression in mathematics is an expression created using variables, constant algebraic numbers, and algebraic operations. A good example of an algebraic expression is 3x2 2xy + c.

An equation that could be used to determine the number of wreaths that Nicole could decorate using one spool of ribbon is:

W = S / R

where W is the number of wreaths, S is the total length of ribbon on the spool, and R is the length of ribbon used per wreath.

Plugging in the given values, we get:

W = 20 / 2.5

Solving for W, we get:

W = 8

So Nicole could decorate 8 wreaths using one spool of ribbon.

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Answer:

do your own work don't ask any question

Answer:

Step-by-step explanation:

I am assuming you mean the area of the shaded figure.

So, let's do it this way.

Let's subtract the total area from the unshaded area:

7 1/4 x 6 = 43 1/2

Area of non-shaded = 18 1/3

Area of shaded = 43 1/2 - 18 1/3 = 25 1/6

Hope that helps! :)

-Aphrodite

Answer:

520 because 65 x 8 is 520

Step-by-step explanation:

Answer:

Solutions given:

given equation is

x²-5x-2=0

comparing above equation with ax²+bx+c,we get

a=1

b=-5

c=-2

By using quadratic equation

x=[tex] \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]

x=[tex] \frac{ 5± \sqrt{ {-5}^{2} - 4*1*-2}}{2*1} [/tex]

x=[tex] \frac{ 5± \sqrt{ 25+8}}{2*1} [/tex]

x=[tex] \frac{ 5± \sqrt{ 33 }}{2} [/tex] is a required answer.

Answer:

[tex]x=\dfrac{5 \pm \sqrt{33}}{2}[/tex]

Step-by-step explanation:

Quadratic Formula

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]

Given equation:

[tex]x^2-5x-2=0[/tex]

Comparing the given equation with [tex]ax^2+bx+c=0[/tex] to find the values of a, b and c:

Substitute these values into the quadratic formula and solve for x:

[tex]\implies x=\dfrac{-(-5) \pm \sqrt{(-5)^2-4(1)(-2)}}{2(1)}[/tex]

[tex]\implies x=\dfrac{5 \pm \sqrt{25+8}}{2}[/tex]

[tex]\implies x=\dfrac{5 \pm \sqrt{33}}{2}[/tex]