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3 years ago

14

25 POINTS!!!! AND BRAINLIEST!

1 answer:

NISA [10]3 years ago

3 0

It would mean for them to. Sooo like One of the basic forms they would use for the internet will help over however many people work there. If median and mode is 25 you would want to set the numbers together

.... hope this helps?

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Which equation represents a line that is perpendicular to line PQ?

valentinak56 [21]

Answer:

B. y=1/3x+4

Step-by-step explanation:

Hi there!

We are given the line PQ and we want to find the line that is perpendicular to it

Perpendicular lines have slopes that are negative and reciprocal. When they are multiplied together, the result is -1

So first, let's find the slope of the line PQ

The point P is given as (-8, 7) and the point Q is given as (-4, -5)

The formula for the slope calculated from two points is $\frac{y_2-y_1}{x_2-x_1}$ where ( $x_{1}$ $y_1$ ) and ( $x_2$ , $y_2$ ) are points

We have the needed information for the slope, but let's label the values of the points to avoid any confusion

x1=-8

y1=7

x2=-4

y2=-5

Now substitute into the formula (m is the slope, and remember: the formula contains SUBTRACTION):

m= $\frac{(-5-7)}{(-4--8)}$

simplify

m= $\frac{-5-7}{-4+8}$

add

m=-12/4

divide

m=-3

So the slope of the line PQ is -3

As said above, perpendicular lines have slopes that have a product of -1

So to find the slope of the line perpendicular to PQ, use this formula:

-3m=-1

divide both sides by -3

m=1/3

The only line that has a slope of 1/3 is B (y=1/3x+4), so B is the answer.

Hope this helps!

3 0

3 years ago

Help!!<br> I can't solve it

Rudiy27

Trick Question their is No 1

7 0

3 years ago

Solve for e: e=mc2 m=3 c=6

Dafna11 [192]

E=3x6x6
e=3x36
e=108
hope it helps

5 0

3 years ago

Sayyy I need help on dis work

LenKa [72]

There is nothing there sorry

4 0

3 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

4 years ago