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Ivahew [28]
3 years ago
10

Which statements are true for the functions g(x) = x^2 and h(x) =-x^2? Check all that apply

Mathematics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

h(x) is the NEGATIVE of g(x)

Step-by-step explanation:

Where are the statements?  Please share them next time.

h(x) is the NEGATIVE of g(x):  h(x) = -x^2 = -g(x)

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Rectangle abcd is similar to rectangle pqrs given that ab =14cm bc =8cm and pq=12cm calculate the length of qr
ohaa [14]

Answer: 6.8571 (Round as needed)

Hope this is correct

Step-by-step explanation:

This is a simple ratio problem

Using the similarity statement we can say 14:12, (ab:pq)

That is our ratio.

So we do 14/12 = 8/x, we solve this using algebra to get our answer.

Hope this is correct.

7 0
2 years ago
Please help !!
xxMikexx [17]
I hope this helps you

5 0
3 years ago
Read 2 more answers
Twice the difference of a number and 8 equals 2
bixtya [17]

Answer:

x=16

Step-by-step explanation:

2(x/8)=2

2x / 16=2

   x 16 x 16

2x=32

2x/2= 32/2

x= 16    

6 0
3 years ago
Show how you got the answer
rewona [7]
Subtract the 45 minutes they were eating from 1:45pm since that’s when they arrived. You now have 1:00pm, you now just need to find out how long it is from 10:20 am to 1:00pm. HINT: it’s 2h 40m
5 0
3 years ago
Let g be the function given by g(x)=limh→0sin(x h)−sinxh. What is the instantaneous rate of change of g with respect to x at x=π
lorasvet [3.4K]

The <em>instantaneous</em> rate of change of <em>g</em> with respect to <em>x</em> at <em>x = π/3</em> is <em>1/2</em>.

<h3>How to determine the instantaneous rate of change of a given function</h3>

The <em>instantaneous</em> rate of change at a given value of x can be found by concept of derivative, which is described below:

g(x) =  \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Where h is the <em>difference</em> rate.

In this question we must find an expression for the <em>instantaneous</em> rate of change of g if f(x) = \sin x and evaluate the resulting expression for x = \frac{\pi}{3}. Then, we have the following procedure below:

g(x) =  \lim_{h \to 0} \frac{\sin (x+h)-\sin x}{h}

g(x) =  \lim_{h \to 0} \frac{\sin x\cdot \cos h +\sin h\cdot \cos x -\sin x}{h}

g(x) =  \lim_{h \to 0} \frac{\sin h}{h}\cdot  \lim_{h \to 0} \cos x

g(x) = \cos x

Now we evaluate g(x) for x = \frac{\pi}{3}:

g\left(\frac{\pi}{3} \right) = \cos \frac{\pi}{3} = \frac{1}{2}

The <em>instantaneous</em> rate of change of <em>g</em> with respect to <em>x</em> at <em>x = π/3</em> is <em>1/2</em>. \blacksquare

To learn more on rates of change, we kindly invite to check this verified question: brainly.com/question/11606037

4 0
2 years ago
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