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3 years ago

What’s the answer? ANSWER THIS NOW PLEASE

Mathematics

2 answers:

hodyreva [135]3 years ago

7 0

Answer:

v=4

Step-by-step explanation:

-23=-8v+9

-23-9=-8v

-32=-8v

-32/-8=v two negatives make a positive

4=v

levacccp [35]3 years ago

5 0

Answer:

v=4

Step-by-step explanation:

-23=3v+9-11v

-23=9-8v

-32=-8v

v=4

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Find a unit vector in the direction of the vector left bracket Start 3 By 1 Matrix 1st Row 1st Column negative 8 2nd Row 1st Col

MrRissso [65]

Answer:

$\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]$

Step-by-step explanation:

We are required to find a unit vector in the direction of:

$\left[\begin{array}{c}-8\\7\\2\end{array}\right]$

Unit Vector, $\hat{a}=\dfrac{\overrightarrow{a}}{|\overrightarrow{a}|}$

The Modulus of $\overrightarrow{a}$ = $\sqrt{(-8)^2+7^2+(-2)^2}=\sqrt{117}$

Therefore, the unit vector of the matrix is given as:

$\left[\begin{array}{c}-\frac{8}{\sqrt{117} } \\\frac{7}{\sqrt{117} }\\\frac{2}{\sqrt{117} }\end{array}\right]$

5 0

3 years ago

What does the variable r represent in the equation 10r+4=8

Kryger [21]

Answer:

The value of variable r in the equation $10r+4=8$ is $\mathbf{r=0.4}$

Step-by-step explanation:

We need to find the value of variable r in the equation $10r+4=8$

Step 1: Write the equation

$10r+4=8$

Step 2: Subtract 4 on both sides

$10r+4-4=8-4\\10r=4$

Step 3: Divide both sides by 4

$\frac{10r}{10}=\frac{4}{10}\\r=0.4$

So, The value of variable r in the equation $10r+4=8$ is $\mathbf{r=0.4}$

3 0

2 years ago

arrange the expressions in the correct sequence to rationalize the denominator of the expression -(2)/(\sqrt(x+y-2)-\sqrt(x+y+2)

cupoosta [38]

We have to rationalize the denominator:
$\frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} } = \\ \frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} }* \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{ \sqrt{x+y-2}+ \sqrt{x+y+2} }= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-(x+y+2)}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-x-y-2}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2} }{-4}= \\ \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{2}$