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defon
3 years ago
5

suppose that f(x)=x^2 and g(x) = -2/3x^2 which statement best compares that graph of g(x) with the graph of f(x)?

Mathematics
1 answer:
vodka [1.7K]3 years ago
6 0

Answer:

f(x) = x^2 , g(x)= -\frac{2}{3}x^2

And we want to compare the two functions.

The minus signs is a reflection around the x axis and the value of 2/3 is a compression of the original function so then the best answer would be:

The graph of g(x) is the graph of f(x) compressed vertically and reflected over the x axis

Step-by-step explanation:

We have the following two function given:

f(x) = x^2 , g(x)= -\frac{2}{3}x^2

And we want to compare the two functions.

The minus signs is a reflection around the x axis and the value of 2/3 is a compression of the original function so then the best answer would be:

d) The graph of g(x) is the graph of f(x) compressed vertically and reflected over the x axis

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There are 400 students, and 100 buses, so number of student in each bus= 400/100

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3 years ago
Which function represents transforming ƒ(x) = 3x with a reflection over the x-axis and a vertical shift of 4 units?
Oxana [17]

Answer:

D. h(x)=-3x+4

Step-by-step explanation:

we are given

f(x)=3x

Firstly, it is reflected over x-axis

we know that for reflection about x-axis , we can replace y as -y

we get

g(x)=-f(x)

g(x)=-3x

now, it is vertical shift up by 4 units

so, we can add 4 to y-value

we get

h(x)=g(x)+4

h(x)=-3x+4

4 0
3 years ago
Find S18 for geometric series given a5=-6 and a2= -48
Ganezh [65]

an = a1r^(n-1)

a5 = a1 r^(5-1)

-6 =a1 r^4


a2 = a1 r^(2-1)

-48 = a1 r


divide

-6 =a1 r^4

----------------    yields   1/8 = r^3      take the cube root  or each side

-48 = a1 r                     1/2 = r


an = a1r^(n-1)

an = a1 (1/2)^ (n-1)

-48 = a1 (1/2) ^1

divide by 1/2

-96 = a1


an = -96 (1/2)^ (n-1)


the sum

Sn = a1[(r^n - 1/(r - 1)]

S18 = -96 [( (1/2) ^17 -1/ (1/2 -1)]

       =-96 [ (1/2) ^ 17 -1 /-1/2]

      = 192 * [-131071/131072]

 approximately -192

     

       

     

3 0
4 years ago
Solve for the given variable:<br><br> A = P (1 + rt) for t
stepan [7]

A = P(1+rt)

A = P + Prt

A - P = Prt

(A - P)/Pr = t

So the answer is:

t = \frac{A-P}{Pr}

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4 years ago
Find dy/dx if y= (x²-3)³cos2x
Over [174]

Answer:

dy/dx = 6x(x²-3)cos2x - 2(x²-3)³sin2x

Step-by-step explanation:

We use the Product Rule:

y= (x²-3)³cos2x

dy/dx =   (x²-3)³ d(cos2x)/dx + cos2x * d ((x²-3)³)/dx

=   (x²-3)³ * -2sin2x + cos2x * 3(x²-3)*2x

= -2sin2x*(x²-3)³ + 6xcos2x*(x²-3)

=  6x(x²-3)cos2x - 2(x²-3)³sin2x

7 0
3 years ago
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