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3 years ago

How do I do vertex in quadratics?How do I graph the quadratic?

Mathematics

1 answer:

nika2105 [10]3 years ago

3 0

What are you trying to figure out

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Complete the equation of the line through (-9,-9) and (-6,0)

Rudik [331]

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis.

According to the data of the statement we have the following points:

$(x_ {1}, y_ {1}): (- 9, -9)\\(x_ {2}, y_ {2}): (- 6,0)$

We found the slope:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0 - (- 9)} {- 6 - (- 9)} = \frac { 9} {- 6 + 9} = \frac {9} {3} = 3$

Thus, the equation is of the form:

$y = 3x + b$

We substitute one of the points and find b:

$0 = 3 (-6) + b\\0 = -18 + b\\b = 18$

Finally, the equation is:

$y = 3x + 18$

Answer:

$y = 3x + 18$

7 0

3 years ago

What curve passes through the point (1 ,2) and has an arc length on the interval [2,6] given by Integral from 2 to 6 StartRo

Georgia [21]

Answer:

Step-by-step explanation:

Given

Length of curve

$L=\int_{2}^{6}\sqrt{1+64x^{-6}}dx$

Length of curve is given by

$L=\int_{a}^{b}\sqrt{1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x}\right )^2}dx$ over interval a to b

comparing two we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=8x^{-3}$

$dy=8x^{-3}dx$

integrating

$\int dy=\int 8x^{-3}dx$

$y=-4x^{-2}+C$

Curve Passes through (1,2)

$1=-4+C$

$C=5$

curve is

$y+\frac{4}{x^2}=5$

3 0

3 years ago

Please Help thank you

CaHeK987 [17]

Answer:

Missing Number: 70,000,000

3 0

3 years ago

Read 2 more answers

What is the slope and y intercept?

natka813 [3]

The y intercept is -2.

4 0

3 years ago

Yuhh get in too ittt

DanielleElmas [232]

Answer:

Step-by-step explanation:

16x-16=4y

4(4x-4)=4y

y=4x-4

Therefore, the correct answer is choice A. Hope this helps!

7 0

3 years ago

Read 2 more answers