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mel-nik [20]
3 years ago
11

Please help!

Mathematics
2 answers:
Ksivusya [100]3 years ago
7 0
Vertical compression would be 1/13. 
so g(x)=(6/13)x + 2
but if that's not right and it's a stretch then it's
g(x)=78x+2
Alex Ar [27]3 years ago
6 0

Answer: The equation of g(x) is g(x)=\frac{6}{13}x+2.

Explanation:

The given equation is,

f(x)=6x

If the graph of function f(x) is stretched by k then g(x) = kf(x) and if the graph of the function f(x) is compressed by k then g(x)=\frac{1}{k} f(x).

If the graph of function f(x) is translated upward by p units then g(x) = f(x)+p and if the graph of the function f(x) is translated downward by p units  then g(x)=f(x)-p.

It is given that the graph of f(x) is transformed into the graph of g(x) by a vertical compression of 13 and a translation of 2 units up.

g(x)=\frac{1}{13}f(x)+2

g(x)=\frac{1}{13}(6x)+2

g(x)=\frac{6}{13}x+2

Therefore, the the equation for g(x) is g(x)=\frac{6}{13}x+2.

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Answer:

30

Step-by-step explanation:

6 0
2 years ago
Find the equation of the line in slope-intercept form containing the points (6, -1) and (-3, 2).
egoroff_w [7]
Slope intercepf is y=mx+b wherem=slope and b= y intercept
slope is found by doing
(y1-y2)/(x1-x2)
points are (6,-1) and (-3,2)
(x,y)
x1=6
y1=-1
x2=-3
y2=2
subsitute
(-1-2)/(6-(-2))=-3/(6+2)=-3/8
slope=-3/8
subsitute
y=-3/8x+b
subsitute and solve for b
(-3,2)
x=-3
y=2
2=-3/8(-3)+b
2=9/8+b
2=16/8
subtract 9/8 from both sides
16/8-9/8=b
7/8=b
y=-3/9x+7/8 is the equation
5 0
3 years ago
Compute the amount of interest earned in the following simple interest problem. A deposit of $1,600 at 6% for 180 days =
Nikitich [7]
6% of $1,600 is = 96 if you multiple it with 180 : 96 × 180 = 17,280
7 0
3 years ago
Read 2 more answers
Which equation represents the graph of the linear function?
blsea [12.9K]
I think the answer is D

Hope it helps
6 0
2 years ago
Two problems here I need solved! I need every step, so please have that with your answers!!
Free_Kalibri [48]
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

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

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
is not in the domain of the given equation.

It is an extraneous solution.

\therefore \: x = - 1
is the only solution.

QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
7 0
3 years ago
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