Center(-3,-4) radius=10 plz help

I need answer to both of the questions please someone!! 20points

diamong [38]

Answer:

Step-by-step explanation:

x+62=90 degree (being perpendicular)

x=90-62

x=28 degree

x+25=90 (being perpendicular)

x=90-25

x=65 degree

8 0

2 years ago

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Find the equation of the line with slope = 6 and passing through (-9,-58). Write your equation in the form y = mx+ b

stira [4]

Answer:

y=6x-4

Step-by-step explanation:

What we know

the slope is 6
our point is (-9,-58)

-9 is x and -58 is y so we have to find the y-intercept y=mx+b is also the formula for this

-58=6(-9)+b

-58=-54+b

-58=54+b - we changed -54 to a positive 54

54-58=-4

y=6x-4

4 0

2 years ago

Please Hurry!!! What is the domain of f(x) = ?

Evgen [1.6K]

Download an app called photomath
it will help you

3 0

2 years ago

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4x+3=-5 helppp please

Goryan [66]

Answer: x = -2

Step-by-step explanation: To solve for x, in the equation you see here, our goal is to get x by itself.

Our first step will be to isolate the term containing x which in this case is 4x. To isolate 4x, we have to get rid of the +3 by subtracting 3 from both sides of the equation. On the left, +3 and -3 cancel out and we're left with 4x. On the right, -5 - 3 simplifies to -8.

Now we have the equation 4x = -8.

To get x by itself, we divide both sides of the equation by 4. On the left, the 4's cancel and on the right, -8 divided by 3 is -2 so x = -2

5 0

3 years ago

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In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the p

masha68 [24]

In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the population continues to change linearly: A.) Find a formula for the moose population, P , in terms of t , the years since 1990. P ( t ) = B.) What does your model predict the moose population to be in 2008?

Answer:

Model predict the moose population to be in 2008 is 1860

Solution:

Let us consider this problem on graph, taking the x axis to be years since 1990 and y on the graph be the number of moose

We have two points on the graph: (2, 4260) and (6, 3660)

Using the slope formula we can find the slope :

$\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{(3660-4260)}{(6-2)}$

m = -150

Now that we have the slope of the line, we can use the point-slope formula to find the equation for the line as follows:-

$\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-4260=-150(x-2)} \\\\ {y-4260=-150 x+300} \\\\ {y=-150 x+4560}\end{array}$

Since the formula that is being sought is supposed to be interms of P and t, we will replace y with P and x with t

P(t) = -150t + 4560

Number of years from 1990 to 2008 is

2008 - 1990 = 18

So, population in 2008 will be

P(18) = -150 (18) + 4560

P(18) = -2700 + 4560

P(18) = 1860

Thus the model to predict the moose population in 2008 is found

4 0

2 years ago