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MAXImum [283]
3 years ago
15

PLEASE HELP TO BE MARKED THE BRAINLIEST

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
5 0
It would be 25%. Hope this helps
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Can someone help me with this question?
andre [41]
The y-intercept (where x = 0) of the equation is at -20.

For every 60 the x runs, the y falls by -10. Use rise over run to determine the slope of the equation.

\frac{-10}{60} = - \frac{1}{6}

Slope-intercept form is determined by the following equation:

y = mx + b

m is your slope, and b is your y-intercept.

Plug the slope and y-intercept into the equation.

y =  -\frac{1}{6}x - 20
4 0
4 years ago
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If x = −4, which number line shows the value of |x|?
Art [367]

Answer:

4 is the absolute value of -4

Step-by-step explanation:

absolute value is basically how many units from 0 the number is.

4 0
3 years ago
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If f(x) = arcsinx, then f' (√3/2)=
Irina-Kira [14]

You can either use the inverse function theorem or compute the general derivative using implicit differentiation. The first method is slightly faster.

The IFT goes like this: if f(x) is invertible and f(a) = b, then finv(b) = a (where "finv" means "inverse of f").

By definition of inverse functions, we have

f(finv(x)) = finv(f(x)) = x

Differentiating both sides of the second equality with respect to x using the chain rule gives

finv'(f(x)) * f'(x) = 1

When x = a, we get

finv'(b) * f'(a) = 1

or

finv'(b) = 1/f'(a)

Now let f(x) = sin(x), which is invertible over the interval -π/2 ≤ x ≤ π/2. In the interval, we have sin(x) = √3/2 when x = π/3. We also have f'(x) = cos(x).

So we take a = π/3 and b = √3/2. Then

arcsin'(√3/2) = 1/cos(π/3) = 1/(1/2) = 2

4 0
3 years ago
F(x) = x +3 What is the slope
nadezda [96]

Answer:

1

Step-by-step explanation:

If there is nothing by the (x) in the equation there is always an invisible 1 next to it. The full equation is f(x)= 1x+3. The answer is 1. Hope that helps!

8 0
3 years ago
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17×7=119. Write the answer to 0.17×7​
Sergeu [11.5K]

Answer:

1.19..... and yes I'm positive that that's the answer

5 0
3 years ago
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