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3 years ago

Given ƒ(x) = −15x − 13, find x when ƒ(x) = −28

Mathematics

1 answer:

kumpel [21]3 years ago

6 0

Answer:

In the picture.

Step-by-step explanation:

Hope that helps.

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Let C be the event that a randomly chosen person went to college. Let J be the event that a randomly chosen person has a job. Id

Verdich [7]

Answer:

the answer is c..................

4 0

2 years ago

what is the y-intercept of the line perpendicular to the line y=-3/4x+5 that includes the point (-3,-3)?

Deffense [45]

Answer:

Step-by-step explanation:

For this question, your friend is the point-slope form.

Point-slope form = $y-y_1 = m(x-x_1)$

The only variables you replace are m = slope, y1 = y of a given point, and x1 = x of a given point.

First, we need to find the slope that is perpendicular to the equation $y=-\frac{3}{4}x+5$

An equation that is perpendicular to another equation have to be negative reciprocals (negative inverse) of each other.

If the slope is 3, then the negative reciprocal of 3 is $-\frac{1}{3}$ .

So, since the given slope is -3/4, the negative reciprocal is $\frac{4}{3}$ .

We will use the negative reciprocal as m since we are trying to find the equation of line that includes (-3, -3) and is perpendicular to the other equation.

We will use point slope form using the (-3, -3) points and negative reciprocal as m.

$y-y_1 = m(x-x_1)\\y-(-3) = \frac{4}{3} (x-(-3))\\y+3=\frac{4}{3}(x+3)$

Now, we need to convert this into slope-intercept form. The reason why we need to do this is because y = mx + b, where b is the y-intercept. To do this, solve for y - or in other words, isolate y..

$y+3=\frac{4}{3}(x+3)\\y+3=\frac{4x}{3}+4\\y+3-3=\frac{4x}{3}+4-3\\y=\frac{4}{3}x+1$

The y-intercept of the line is 1.

3 0

3 years ago

An object is launched from a platform.

tankabanditka [31]

Answer:

Step-by-step explanation:

Ground level is where h = 0, so solve the equation ...

h(x) = 0

-5(x -4)^2 +180 = 0 . . . . substitute for h(x)

(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36

x -4 = 6 . . . . . . . . . . . . . . positive square root*

x = 10 . . . . . . add 4

The object will hit the ground 10 seconds after launch.

_____

* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.