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3 years ago

I need help please i

Mathematics

2 answers:

KIM [24]3 years ago

8 0

Answer:

Step-by-step explanation:

- 1.2 to 11.2

Rashid [163]3 years ago

4 0

Answer:

Step-by-step explanation:

the answer is somewhere by 5

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How many one inch cubes can fit in a cube of 180 cubic inches

My name is Ann [436]

1x 1 inch cube has volume 1in³, so 180 of these cubes can fit in a 180in³ space.

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3 years ago

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A sphere has a volume of 288 pi in^3. Find the surface area, Express your answer in pi.

Tpy6a [65]

Sphere Volume = 4/3 • π • r³
r^3 = sphere volume / (4/3 * PI)
r^3 = 288*PI / (4/3 * PI)
r^3 = 288 / (4/3)
r^3 = 216
r = 6 inches

Sphere Surface Area = 4 • π • r²

Sphere Surface Area = 4 * PI * 36

Sphere Surface Area = 452.389 square inches

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3 years ago

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How was his payment

dmitriy555 [2]

Subtract off the down payment: 880 - 250 = 630

So he'll pay 630 over 8 equal payments
Divide: 630/8 = 78.75

His monthly payment is $78.75

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3 years ago

Find the solution to this system of equations 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$

Hence,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1$

Substitute it into the second equation:

$23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0$

Substitute them into the first equation:

$3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2$

The solution is $x=2,\ y=0,\ z=-1.$

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4 years ago

Hi can anybody tell me the answer

MArishka [77]

Answer:

8x^5y^7

Step by step explanation

$4 {x}^{2} {y}^{3} \times 2 {x}^{3} {y}^{4} \\ 4 \times 2 \times {x}^{2} \times {x}^{3} \times {y}^{3} \times y ^{4} \\ 8 \times {x}^{2 + 3} \times {y}^{3 + 4} \\ =8 {x}^{5} {y}^{7}$

6 0

3 years ago

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