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Reika [66]
2 years ago
14

Which statement about the relationship between the graph f and the graph of g(x)=7x^2 is true

Mathematics
2 answers:
ololo11 [35]2 years ago
5 0

Answer:

where is F

Step-by-step explanation:

Leno4ka [110]2 years ago
5 0
It's ffffffffffff:))))
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Need some help with the answer to the question
kicyunya [14]

Answer:

(- 4, 27 )

Step-by-step explanation:

Equate the right sides of both equations, that is

x² - 2x + 3 = - 2x + 19 ← subtract - 2x + 19 from both sides

x² - 16 = 0 ← in standard form

(x - 4)(x + 4) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x + 4 = 0 ⇒ x = - 4

Substitute these values into f(x) = - 2x + 19

f(4) = - 2(4) + 19 = - 8 + 19 = 11 ⇒ (4, 11 )

f(- 4) = - 2(- 4) + 19 = 8 + 19 = 27 ⇒ (- 4, 27 )

8 0
3 years ago
Please help me answer these 2 questions with a FULL explanation so I know how to do the rest myself. Giving brainliest.
sveticcg [70]

Answer to 1:

\frac{x^3}{2y^25}

First thing you want to do is combine like terms on the bottom. Since it's multiplication, all you have to do is add the exponents. So for x you'd get x raised to the -9 power and for y you'd get 2y raised to the power of 7

Second thing is the numerator of the fraction. Since this is raising what is inside the paranthesis to a power, you must multiply this power by the powers of x and y, respectively. You should get x raised to the power of -6 and y raised to the power of -18

Here's where it gets complicated. to rationalize the problem, you must add parts from the numerator and denominator or vice versa. You cannot have any negative exponents anywhere in the fraction.  So that means for x you have to add 9 to the x on the top to cancel out the -6 and rationalize the remaining x's. You should end up with x^3.

For y, you work the opposite way. Since the negatives are only on the top, you simply add 18 to the bottom y's and it remains rationalized.

If done correctly this should come out to x^3/2y^25

Answer to 2:

(x^20y^12)/256

First thing you can do on this one is to cancel out the x^0s. Anything to the power of 0 equals one so only the coefficients will remain. Multiply those together and you'll get 4.

Next you will rationalize the ys. Add the bottom 3 to the top to cancel out the denominator and you're left with y^-3. Leave it like this for now.

Now is the difficult part. You must take everything that is inside the parenthesis to the power of -4. That includes the coefficients and the exponents. 4^-4 = 1/256. This means that you now must move the coefficients to the bottom. You should currently have y^-3/(256x^5).

Now take the exponents to the power using the same rules as question 1. -3*-4 = 12 and 5*-4 = -20. You should now have y^12/(256x^-20)

Lastly, we must rationalize the denominator. To do so, move the x^-20 to the numerator and make the exponent positive. After doing this, you have the answer: (x^20y^12)/256

3 0
2 years ago
Who should fill out the W-2 form?
elena-14-01-66 [18.8K]
It’s A THE ANSWER IS A
7 0
3 years ago
Read 2 more answers
You have four quiz grades. Quiz 1: 17/30, Quiz 2: 17/25, Quiz 3: 38/50, Quiz 4: 18/25 What is your ordinary quiz average?
Komok [63]

Answer:

.7 or 70%

Step-by-step explanation:

17/30 + 17/25 + 38/50 + 18/25 = 2.727

2.727 divided by 4 (the amount of addends) = .7

5 0
3 years ago
Read 2 more answers
Which of the following equations could be th equation to represent the given graph? Make sure you explain your answer thoroughly
Alex17521 [72]

Answer:

Option (C) is correct.

Step-by-step explanation:

The given options of the possible equation for the graph are as follows:

(A) y=2\left(\frac{3}{2}\right)^x \\\\(B) y=-2\left(\frac{3}{2}\right)^{-x} \\\\(C) y=2\left(\frac{2}{3}\right)^x \\\\(D) y=-2\left(\frac{2}{3}\right)^{-x} \\\\

The given graph is decreasing and at x=0, y=2.

So, first checking the value of the given options for x=0

(A) y=2\left(\frac{3}{2}\right)^0=2\times 1= 2 \\\\(B) y=--2\left(\frac{3}{2}\right)^{-0}= -2\times 1= -2 \; (not\; possible) \\\\(C) y=2\left(\frac{2}{3}\right)^0= 2\times 1= 2 \\\\(D) y=2\left(\frac{2}{3}\right)^{-0} = -2\times 1= -2 \; (not\; possible)

As, for x=0, y=2, so options (C) and (D) are not possible, so rejected.

Now, checking the nature (increasing or decreasing) of the given equation by differentiating it.

For option (A),

\frac{dy}{dx}=2\left(\frac{3}{2}\right)^{x}\times \ln\left(\frac{3}{2}\right)

As \ln \left(\frac{3}{2}\right)=\ln(1.5)>0 \;and\; \left(\frac{3}{2}\right)^{x} >0

So, \frac{dy}{dx}>0

Therefore, the function in option (A) is increasing function.

Similarly, for option (C),

\frac{dy}{dx}=2\left(\frac{2}{3}\right)^{x}\times \ln\left(\frac{2}{3}\right)

As \ln \left(\frac{2}{3}\right)=\ln(0.67)0

So, \frac{dy}{dx}

Therefore, the function in option (C) is decreasing function.

As the given graph is decreasing, so, (C)  representsy=2\left(\frac{2}{3}\right)^x the given graph.

Hence, option (C) is correct.

6 0
3 years ago
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