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3 years ago

25 POINTS PLEASE HELP SOMEONE WHO CAN

Mathematics

2 answers:

NeTakaya3 years ago

8 0

The answer I got is (3,6, -1)

liberstina [14]3 years ago

8 0

Y = 6
x = 3
z = -1
use gaussian elimination

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Which is a valid prediction about the continuous function

RoseWind [281]

Answer:

The answer is A

Step-by-step explanation: yesmgmhmhwibdcibfeiuebuidwbicncocui bcobcoucb9ucxbicb ccuhs ixb cuxbcubci c I took the test oundfiubf9uebaocbxidcb du9bedbc9ubci ciucbicb9cubc9 cibcu cybce8yeb9ceb9ueboicnococbjkdsbcs djcocjcnocjdsndcn cdijcijbcdijbcbdsi cijbsdhvhvchi

3 0

3 years ago

Solve each system by substitution. 1) y = 7x - 10 y=-3

djyliett [7]

-3=7x-10. -7x=3-10. X=1. You can check the answer by plugging in the numbers.

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

Can someone please help me with this

ASHA 777 [7]

X is between -6 and 11 and y between -6 and 8

3 0

3 years ago

Read 2 more answers

What is the solution of the system X - 3y = -13 and 5x + 7y = 34

hammer [34]

Answer:

The solution for this system is x = 1/2 and y = 9/2

Step-by-step explanation:

1. We have this system of equations:

x - 3y = -13 and 5x + 7y = 34

2. Let's try to find out the value of x on the first equation:

x - 3y = - 13

x = 3y - 13

3. Now, let's replace x on the second equation for finding y:

5x + 7y = 34

5 (3y - 13) + 7y = 34

15y - 65 + 7y = 34

22y = 34 +65

22y = 99

2y = 9 (Dividing by 11 at both sides)

y = 9/2 (Diving by 2 at both sides)

4. Now, let's find out the value of x on the first equation, using the value of y:

x - 3y = - 13

x - 3 (9/2) = - 13

x - 27/2 = -13

x = -13 + 27/2

x = (-26 + 27)/2 (Adding - 13 and 27/2)

x = 1/2

5. Finally, let's prove the values of x and y are correct on the second equation of the system:

5x + 7y = 34

5 (1/2) + 7 (9/2) = 34

5/2 + 63/2 = 34

(5 + 63)/2 = 34

68/2 = 34

34 = 34

It has been proved that x = 1/2 and y = 9/2 are correct

5 0

4 years ago