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Kaylis [27]
3 years ago
5

Plzzzzzzzz help with your own handwritten picture 120−96×[5 ÷ {23 − 3 × (12 − 22 − 16)}]

Mathematics
1 answer:
laiz [17]3 years ago
8 0
This is what I got Hope the helped!!

Answer:

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MaRussiya [10]

set them equal to their sum. which in this case is 141 then solve for x and plug it into your original equation to get your answer.

3 0
3 years ago
A certain dodgeball court is a circle with a square perfectly inscribed inside it. the square represents the playing field, whil
cluponka [151]
<span>Exactly 8*pi - 16
 Approximately 9.132741229 For this problem, we need to subtract the area of the square from the area of the circle. In order to get the area of the circle, we need to calculate its radius, which will be half of its diameter. And the diameter will be the length of the diagonal for the square. And since the area of the square is 16, that means that each side has a length of 4. And the Pythagorean theorem will allow us to easily calculate the diagonal. So: sqrt(4^2 + 4^2) = sqrt(16 + 16) = sqrt(32) = 4*sqrt(2) Therefore the radius of the circle is 2*sqrt(2). And the area of the circle is pi*r^2 = pi*(2*sqrt(2)) = pi*8 So the area of the rest areas is exactly 8*pi - 16, or approximately 9.132741229</span>
6 0
3 years ago
Which statement describes the graph of f(x)=-4x^3-28x^2-32x+64
lord [1]
The only that has the shape of the one below.

y = -4(x^3 + 7x^2+8x - 16)
Let x = 1
y = - 4(1 + 7 + 8 - 16)
y = 0

y = -4(x -1)(x^2 + 8x + 16)
y = -4(x - 1)(x + 4)^2


7 0
4 years ago
Read 2 more answers
The radius of the base of the cone is 5 cm.
makkiz [27]

4/2 3.14
or something
4 0
3 years ago
A flat rectangular piece of aluminum has a perimeter of 62 inches. The length is 15
Salsk061 [2.6K]

Answer:

So, the required width of rectangular piece of aluminium is 8 inches

Step-by-step explanation:

We are given:

Perimeter of rectangular piece of aluminium = 62 inches

Let width of rectangular piece of aluminium = w

and length of rectangular piece of aluminium  = w+15

We need to find width i.e value of x

The formula for finding perimeter of rectangle is: Perimeter=2(Length+Width)\\

Now, Putting values in formula for finding Width w:

Perimeter=2(Length+Width)\\\\62=2(w+15+w)\\62=2(2w+15)\\62=4w+30\\62-30=4w\\4w=32\\w=\frac{32}{4}\\w=8

After solving we get the width of rectangular piece :w = 8

So, the required width of rectangular piece of aluminium is 8 inches

6 0
3 years ago
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