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2 years ago

Write the equation for the line that goes through point (12, 15) with slope = −2.

Mathematics

2 answers:

pogonyaev2 years ago

7 0

Answer:

y= -2x +39

Step-by-step explanation:

y= mx+b

m is the slope which is given -2

since we don't know b we can solve the rest using the formula y-y1{ m(x-X1(

y - 15 = -2(x-12)

simplify

y-15= -2x +24

y= -2x + 39

Alexeev081 [22]2 years ago

7 0

Y=-2x+39 is what i got

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PLS HELP! WILL GIVE BRAINLIEST

Leviafan [203]

Answer:

The center (2, 2.5), radius $\sqrt{37}$ / 2

Step-by-step explanation:

The standard form of the equation of a circle is:

( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius

Rewrite the given equation in the standard form:

2x^2 + 2y^2 - 8x + 10y + 2 = 0
x^2 - 4x + y^2 + 5y = -1
x^2 - 4x + 2^2 + y^2 + 5y + (5/2)^2 = -1 + 4 + 25/4
(x - 2)^2 + (y + 2.5)^2 = 37/4

The center is:

(2, 2.5)

And radius is:

$\sqrt{37}$ / 2

4 0

2 years ago

Could someone help me rnnn?

GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola: $y=3x^2-4$

Step-by-step explanation:

Given:

y-intercept of parabola: -4
parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola: $y=a(x-h)^2+k$

(where (h, k) is the vertex and $a$ is some constant)

Substitute point (0, -4) into the equation:

$\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}$

Substitute point (-2, 8) and $ah^2+k=-4$ into the equation:

$\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}$

Substitute point (1, -1) and $ah^2+k=-4$ into the equation:

$\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}$

Equate to find h:

$\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}$

Substitute found value of h into one of the equations to find a:

$\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}$

Substitute found values of h and a to find k:

$\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}$

Therefore, the equation of the parabola in vertex form is:

$\implies y=3(x-0)^2-4=3x^2-4$

So the vertex of the parabola is (0, -4)

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