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4 years ago

Answer asap please i don’t understand

Mathematics

1 answer:

Anna35 [415]4 years ago

6 0

Answer:

11x+2y+21 x = 15

2x+3y+6z = -29

13x+9x+2x =22

Step-by-step explanation:

This is an augmented matrix

It is the coefficients of the variables and the last number is the constant

11x+2y+21 x = 15

2x+3y+6z = -29

13x+9x+2x =22

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Solve for X!!! i neeed help <br> spam will be reported.

NNADVOKAT [17]

Answer:

x=5

Step-by-step explanation:

All we have to do for this equation is cross multiply. This means we multiply 7 and 5 together and 5 and x+2 together. This will make the equation 35=5(x+2). Simplify this to get 35=5x+10. Subtract 10 from both sides to get 5x=25. You want to isolate x so you have to divide both sides by 5. This will make the equation x=5.

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4 0

3 years ago

Solve -39 4/9 - (-10 2/9) - (18 2/9)

tatyana61 [14]

12850/27
Sorry that’s a really big number

7 0

3 years ago

Usimov [2.4K]

Answer:

Yes, 19.3649167 < 20

Step-by-step explanation:

Do the Pythagorean theorem to figure out the height.

35^2 + x^2 = 40^2

1225 + x^2 = 1600

x^2 = 375

x = 19.3649167

7 0

3 years ago

A. m∠ADB = (7x + 27)° Find x .

Helen [10]

Answer:

x = 9

y = 12

Step-by-step explanation:

✔️m<ADB = 90°

7x + 27 = 90

7x = 90 - 27

7x = 63

Divide both sides by 7

x = 9

✔️BC = 2(BD)

3y - 7 = 2(y + 3)

3y - 7 = 2y + 6

Collect like terms

3y - 2y = 7 + 6

y = 12

8 0

3 years ago

Match each spherical volume to the largest cross sectional area of that sphere

zlopas [31]

Answer:

Part 1) $324\pi\ units^{2}$ ------> $7,776\pi\ units^{3}$

Part 2) $36\pi\ units^{2}$ ------> $288\pi\ units^{3}$

Part 3) $81\pi\ units^{2}$ ------> $972\pi\ units^{3}$

Part 4) $144\pi\ units^{2}$ ------> $2,304\pi\ units^{3}$

Step-by-step explanation:

we know that

The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere

Part 1) we have

$A=324\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$324\pi=\pi r^{2}$

Solve for r

$r^{2}=324$

$r=18\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=18\ units$

substitute

$V=\frac{4}{3}\pi (18)^{3}$

$V=7,776\pi\ units^{3}$

Part 2) we have

$A=36\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$36\pi=\pi r^{2}$

Solve for r

$r^{2}=36$

$r=6\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=6\ units$

substitute

$V=\frac{4}{3}\pi (6)^{3}$

$V=288\pi\ units^{3}$

Part 3) we have

$A=81\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$81\pi=\pi r^{2}$

Solve for r

$r^{2}=81$

$r=9\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=9\ units$

substitute

$V=\frac{4}{3}\pi (9)^{3}$

$V=972\pi\ units^{3}$

Part 4) we have

$A=144\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$144\pi=\pi r^{2}$

Solve for r

$r^{2}=144$

$r=12\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=12\ units$

substitute

$V=\frac{4}{3}\pi (12)^{3}$

$V=2,304\pi\ units^{3}$

5 0

3 years ago

Read 2 more answers