All categories

3 years ago

Find the equation of the line specified.

1 answer:

7 0

The point slope form of a straight line is y-y1 = m(x-x1)

y -(-3) = -7(x-5)

y + 3 = -7(x-5)

y = -7x + 35 - 3

y = -7x + 32

You might be interested in

F(x) = x2. What is g(x)?

puteri [66]

Answer:

C: g(x) = 3x^2

Step-by-step explanation:

Note that (1, 1) is on the graph of f(x) and that (1, 3) is on the graph of g(x). This indicates that the coefficient of x^2 is 3 for g(x).

5 0

3 years ago

Please help me I will give you brainlest!!!

Svetllana [295]

Answer:

I am just trying.

Step-by-step explanation:

MLON=HIJK....

To tell the honest truth I dont have a clue. Wait for someone else to answer.

6 0

3 years ago

What is the distance between (-4,0)and (-4,-5)

Nastasia [14]

Answer:

<h2><em><u>ᎪꪀsωꫀᏒ</u></em></h2>

5 units

Step-by-step explanation:

$\sf{}$

4 0

3 years ago

Does anyone know the answer to this

algol [13]

Answer:

x² + (y+3)² = 4²

Step-by-step explanation:

A close look at the diagram reveals that the center of this circle is (0, -3) and that the radius is 4 units.

Starting from the general equation of a circle, (x - h)² + (y - k)² = r²

and substituting the givens (h = 0, k = -3, r = 4), we get:

x² + (y+3)² = 4². This is Answer B.

4 0

3 years ago

Use the equation to answer the following question y=(x-3(x+2)/(x+4)(x-4)(x+2)

givi [52]

Answer:

See below

Step-by-step explanation:

The given rational function is;

$y=\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}$

The given function is not continuous where the denominator is equal to zero.

$(x+4)(x-4)(x+2)=0$

The function is discontinuous at $x=-4,x=4,x=-2$

b) The point at x=-2 is a removable discontinuity(hole) because (x+2) is common to both the numerator and the denominator.

The point at x=-4 and x=4 are non-removable discontinuities(vertical asymptotes)

c) The equation of the vertical asymptotes are x=-4 and x=4

To find the equation of the horizontal asymptote, we take limit to infinity.

$\lim_{x\to \infty}\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}=0$

The horizontal asymptote is y=0

8 0

3 years ago