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qaws [65]
2 years ago
13

Simplify the product. Write your answer in standard notation 2y(5y^2 + 3)

Mathematics
1 answer:
GarryVolchara [31]2 years ago
3 0

Answer:

10y^3+6y

Step-by-step explanation:

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Factor 6x4 – 5x2 + 12x2 – 10 by grouping. What is the resulting expression?
VARVARA [1.3K]
I hope this helps you

8 0
3 years ago
Read 2 more answers
What is the value of P(9.1)? What does it tell us about the side length and perimeter
nadya68 [22]

Answer:

the answer is 36.4

Step-by-step explanation:

you have to multiply it by 4 so it it will look like this 9.1*4=36.4

4 0
3 years ago
A water tank is in the shape of a right circular Austin simulato water tank is in the shape of a right circular Cylinder with a
Burka [1]
⓵ To calculate the volume of a right circular cylinder, the formula is π times the radius of the circular base² time the height of the cylinder.

⓶ Now that we know that the equation to calculate the volume of a right circular cylinder is :

V = π x (r²) x h

You need to find the numbers to replace the volume (V) and the height (h) in the formula.

We already know that the volume is 320 square feet and that the height is 20 feet.

So we are left with a formula looking like this :

320 = π x (r²) x 20

⓷ Now we need to find the radius of the circular base! To do so, you need to solve this equation and isolate the “r”. Start by simplifying the right side :

320 = π x (r²) x 20
÷20 ÷20

↓

16 = π x r²
÷π ÷π

↓

5,09 ⋍ r²
√ √

↓

2,26 feet ⋍ r

⓸ Now that we knoe the value of the radius of the circular base, all there’s left to do is multiply this number by two in order to find the diameter of the water tank :

2,26 x 2 = d

↓

4,51 feet ⋍ d

So your final answer is : the diameter of the water tank is about 4,51 feet.

** Since I devided by “π”, all the answers I wrote from that point are rounded to the nearest hundredths just to make things easier to visualize, but I kept all of the decimals when doing the calculations. So it is possible that your answer might differ slightly from mine if you use the rounded numbers to calculate everything. Just keep that in mind!

I hope this helped, if there’s anything just let me know! ☻
4 0
3 years ago
A rectangle ahs a long side of a length 8.5 cm its perimeter is 24cm work out the length of the shorter side
Phantasy [73]

Answer:

Width (length of shorter side) = 3.5 cm.

Step-by-step explanation:

L = 8.5 cm

perimeter, P = 24 cm

2(L+W) = 24

substitute L = 8.5

2(8.5+W) = 24

8.5 + W = 24/2 = 12

W = 12 - 8.5 = 3.5 cm

5 0
3 years ago
Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the
dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

I = \int \int_E \int zdV where E is bounded by the cylinder y^2 + z^2 = 81 and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and y^2 + z^2 = 81

∴ z^2 = 81 - y^2

z = \sqrt{81 - y^2}

Thus, z lies between 0 to \sqrt{81 - y^2}

GIven curve x = 0 and y = 9x

x =\dfrac{y}{9}

As such,x lies between 0 to \dfrac{y}{9}

Given curve x = 0 , x =\dfrac{y}{9} and z = 0, y^2 + z^2 = 81

y = 0 and

y^2 = 81 \\ \\ y = \sqrt{81}  \\ \\  y = 9

∴ y lies between 0 and 9

Then I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix}    ^ {\sqrt {{81-y^2}}}_{0} \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{(\sqrt{81 -y^2})^2 }{2}-0  \end {bmatrix}     \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{{81 -y^2} }{2} \end {bmatrix}     \ dxdy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0}    \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix}     \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81 \  y -y^3} }{18} \end {bmatrix}     \ dy

I = \dfrac{1}{18} \int^9_{y=0}  \begin {bmatrix}  {81 \  y -y^3}  \end {bmatrix}     \ dy

I = \dfrac{1}{18}  \begin {bmatrix}  {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}}  \end {bmatrix}^9_0

I = \dfrac{1}{18}  \begin {bmatrix}  {40.5 \ (9^2) - \dfrac{9^4}{4}}  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  3280.5 - 1640.25  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  1640.25  \end {bmatrix}

I = 91.125

4 0
3 years ago
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