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3 years ago

The value of y varies directly with x. If x = 6, then y = 72. What is the value of x when y = 108 ?

Mathematics

1 answer:

lana66690 [7]3 years ago

5 0

Answer: The Answer is: 9

Step-by-step explanation:

I dont know what type of level this is but I got it some how. If its Pre-Algerba you are good cause I am in that math.

What you will need to do is propotions

6/x = 72/108

You have to cross mulitply to find the value of x

Then you would have to multiply 108 x 6= 648

After you would have to cross multiply 72x

You divide 648= 72x and you will get your answer which is 9.

I hope this helped. :)

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Based on the pattern, which power of 10 would have a value of StartFraction 1 Over 1,000,000 EndFraction?

Hunter-Best [27]

Answer:

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Step-by-step explanation:

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2 years ago

There are 27 pupils. There are twice as many boys than girls. how many boys and girls are there

Anna71 [15]

Let x = the numbers of girls
thus
the numbers of boys = 2x
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3 years ago

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Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,

Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j} = {(1,0), (0,1)}

is a base in $\mathbb{R}^2$

and F is linear, then

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

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3 0

3 years ago

Which factor do 64x^2+48x+9 and 64x^2 - 9 have in common?

erastovalidia [21]

<span>(8x +3) (8x-3)
</span>share the same terms

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3 years ago

Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

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3 years ago