Help!! What's the answer for his problem?

Mathematics

2 answers:

meriva3 years ago

4 0

Break it into three sections, then solve for the respective volumes of the three boxes. Your answer is 200cm^3

pashok25 [27]3 years ago

3 0

The answer is 200cm^3

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THIS IS PART B So plz right answer

SIZIF [17.4K]

Answer:

that is confusing

Step-by-step explanation:

6 0

2 years ago

10) g(n) = -21+ 3n; Find gl-6) A) 152 B) -20 C) 1 D) 90

defon

10) g(n) = -21+ 3n; Find gl-6)
A) 152
B) -20
C) 1
D) 90
Your answer would be B

If this helped can I please get brainliest

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2 years ago

Write the quadratic function f(x) = 3x2 - 6x + 9 in vertex form.

Y_Kistochka [10]

The answer is y=3(x-1)^2+6

7 0

2 years ago

Read 2 more answers

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.

7 0

3 years ago

Find the mean absolute deviation for the set below s equals 65 , 90, 85, 70, 70, 95, 55

ddd [48]

75.7 I think that is the answer

5 0

3 years ago