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3 years ago

Write an equation in slope-intercept form for the line passing through the pair of points.

Mathematics

1 answer:

Umnica [9.8K]3 years ago

8 0

First find your slope! by doing
y2-y1/x2-x1
in this case it would be
2-(-3)/-1-(-6)
It would give you 5/5 which would reduce to 1 meaning your slope equals 1
with your slope now you should use the point slope formula y-y1=m(x-x1)

i’m going to use the point (-1,2) so the equation would be

y-2=1(x-(-1))

simplifying that you would get

y-2=1x+1

then put it in y=mx+b form and get

y=x+3

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tatuchka [14]

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Step-by-step explanation:

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3 years ago

The probability that a farmer is in debt is 0.700. The probability that a farmer is in debt and also lives in the Midwest is 0.2

bogdanovich [222]

Answer:

0.4

Step-by-step explanation:

Given that:

P(debt) = P(D) = Probability of being in debt = 0.7

P(debt n Midwest) = P(Dn M) = probability of being in debt and lives in Midwest = 0.280

The probability that a randomly selected farmer lives in the Midwest given that he is in debt is?

P(M | D) = p(D n M) / p(D)

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P(M | D) = 0.4

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3 years ago

List the next four multiples of each fraction 1/4

julsineya [31]

Answer:

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Step-by-step explanation:

1/2

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3 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72

$x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37$

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0

3 years ago

Whats the vertex at of y-13=2(x+4)^2

Kipish [7]

The vertex form:
$y=a(x-h)^2+k$
(h,k) - the coordinates of the vertex

$y-13=2(x+4)^2 \\
y=2(x+4)^2+13 \\
y=2(x-(-4))^2+13 \\ \\
h=-4 \\ k=13$

The coordinates of the vertex are (-4,13).

7 0

3 years ago

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