Find the volume of the composite figure. Drag and drop the correct numbers into the boxes to complete the answers. Please help!!

Mathematics

1 answer:

garri49 [273]3 years ago

5 0

Answer:

Rectangular prism: 308

Triangular prism: 98

Composite figure: 406

Step-by-step explanation:

4 x 7 x 11 = 308

(4x7/2) x 7 = 98

308 + 98 = 406

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Use pascal's triangle to expand the binomial (d-5y)^6 please show work

Ratling [72]

Pascals triangle to the 6th:
1 x^0
1 1 x^1
1 2 1 x^2
1 3 3 1 x^3
1 4 6 4 1 x^4
1 5 10 10 5 1 x^5
1 6 15 20 15 6 1 x^6

the problem is to the 6th power so your going to use the 6th row of pascals triangle (don't count the first row). these numbers represent the coefficients of the variables

1(d-5y)^6 + 6(d-5y)^5 + 15(d-5y)^4 + 20(d-5y)^3 + 15(d-5y)^2 + 6(d-5y) + 1
then simplify

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3 years ago

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Each lap around a park is 0.58 miles. Jessica walked 3.25 laps around the park. How many miles did Jessica walk around the park?

xz_007 [3.2K]

0.58x3.25=1.885
c.)1.885 miles

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3 years ago

What are the equivalent fractions with a common denominator for 2/5 and 1/2?

Svetlanka [38]

So 1/2 is 5/10 and 2/5 is 4/10 common denonominator is 10 ( i think )

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3 years ago

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6x - 2y = 26 2x + 3y = 5 Solve for X and Y

Softa [21]

Answer:

y is equal to -1

x is equal to 4

Step-by-step explanation:

The first thing you would do is multiply the bottom equation by a negative 3. You will get - 6x - 9y = -15. Keep the top equation the same. Both of the 6x's cancel. You then get -11y = 11 so y is equal to negative 1.

Since you have Y, you can now plug that in to any equation to find the value of x. -6x - 2(-1) = 26. You're left with 4. Plug both of the values into each equation to double check.

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3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

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3 years ago