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3 years ago

Write an equation in point-slope form.

1 answer:

5 0

First: let (8,3) be the coordinates of the point A
so equation of straight: y-yA=slope(x-xA)
y-3=6(x-8)
second: y-yB=slope(x-xB)
y-(-5)=-2/5(x-(-3))
y+5=-2/5(x+3)

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What are the steps to solving this equation. 2(x+1)-4=4

puteri [66]

1.Muliply parenthesis by 2

2x+2-4=4

2.calculate the difference

2x-2=4

3.move constant to the right side and change its sign

2x=4+2

4.add the numbers

2x=6

5. Divide both sides of the equation by 2

x=3

3 0

4 years ago

Please help, algebra is annoying

padilas [110]

The zeros are the x-intercepts, which are -2 and 4.

The axis of symmetry is x = 1

The vertex of the graph is (1,9)

7 0

3 years ago

Match each function to its corresponding sequence.

givi [52]

The sequence and their functions are:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....
f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....
f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....

<h3>How to match the functions?</h3>

To do this, we simply set values for n and calculate the function values using the current value of n.

So, we have:

Function 1: f(n) = f(n - 1) + 6n - 5 where f(1) = 3

Let n = 2

f(2) = f(1) + 6(2) - 5 = 3 + 12 - 5 = 10

Let n = 3

f(3) = f(2) + 6(3) - 5 = 10 + 18 - 5 = 23

None of the sequence follows the pattern 3, 10, 23....

Function 2: f(n) = f(n - 1) + 2n - 1 where f(1) = 3

Let n = 2

f(2) = f(1) + 2(2) - 1 = 3 + 4 - 1 = 6

Let n = 3

f(3) = f(2) + 2(3) - 1 = 6 + 6 - 1 = 11

None of the sequence follows the pattern 3, 6, 11....

Function 3: f(n) = f(n - 1) + 6 where f(1) = 3

Let n = 2

f(2) = f(1) + 6 = 3 + 6 = 9

Let n = 3

f(3) = f(2) + 6 = 9 + 6 = 15

None of the sequence follows the pattern 3, 9, 15....

Function 4: f(n) = f(n - 1) + 7 where f(1) = 3

Let n = 2

f(2) = f(1) + 7 = 3 + 7 = 10

Let n = 3, 4 and 5

f(3) = f(2) + 7 = 10 + 7 = 17

f(4) = f(3) + 7 = 17 + 7 = 24

f(5) = f(4) + 7 = 24 + 7 = 31

So, we have:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....

Function 5: f(n) = f(n - 1) + 4n - 2 where f(1) = 3

Let n = 2

f(2) = f(1) + 4(2) - 2 = 3 + 8 - 2 = 9

Let n = 3, 4 and 5

f(3) = f(2) + 4(3) - 2 = 9 + 12 - 2 = 19

f(4) = f(3) + 4(4) - 2 = 19 + 16 - 2 = 33

f(5) = f(4) + 4(5) - 2 = 33 + 20 - 2 = 51

So, we have:

f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....

Function 6: f(n) = 2[f(n - 1)] where f(1) = 3

Let n = 2

f(2) = 2 * f(1) = 2 * 3 = 6

Let n = 3, 4 and 5

f(3) = 2 * f(2) = 2 * 6 = 12

f(4) = 2 * f(3) = 2 * 12 = 24

f(5) = 2 * f(4) = 2 * 24 = 48

So, we have:

f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....

Read more about sequence at:

brainly.com/question/6561461

#SPJ1

7 0

2 years ago

Another one omg this is annoying

Alexeev081 [22]

Answer:

Correct option: first one

Step-by-step explanation:

The equation f(x) = 5x + 1 is a function, because each value of x gives only one value of y.

Now let's find the inverse by switching f(x) by x and x by f'(x), then isolating f'(x):

$x = 5f'(x) + 1$

$5f'(x) = x - 1$

$f'(x) = \frac{x - 1}{5}$

The inverse f'(x) is also a function, because each value of x gives only one value of x.

So we have that both the equation and its inverse are function, therefore the correct answer is the first option.

7 0

3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

8 0

3 years ago