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Yuki888 [10]
3 years ago
11

Given: f(x) = x - 7 and h(x) = 2x + 3

Mathematics
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

f(h(x)) = 2x - 4

Step-by-step explanation:

Substitute x = h(x) into f(x), that is

f(h(x))

= f(2x + 3)

= 2x + 3 - 7

= 2x - 4

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The diameter of the circle is 9.7 m . Calculate the length of its circumference. Round to the nearest tenth and remember to use
RSB [31]

Answer:

30.4m

Step-by-step explanation:

Diameter(D) = 2r (radius)

D = 9.7m

r = 9.7m / 2

= 4.85m

C = 2πr

= 2π × 4.85m

= 30.47344873982099m

rounding - Leave it as is if the digit in the tenth is 4 or lower

- round it up if it's 5 or higher.

~~ 30.4m

5 0
2 years ago
What are the things circled called? You can answer the question fully too if you’d like. I really need help in understanding thi
nikitadnepr [17]
They are called the x and y axis
8 0
3 years ago
Find the value of the integral that converges.<br> ∫^-5_-[infinity] x^-2 dx.
Bingel [31]

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

8 0
3 years ago
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olchik [2.2K]
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5 0
3 years ago
The difference between the acute angles of a right angled triangle is 72degree find them .
MaRussiya [10]

The 2 angles add up to 90 degrees so

x + y = 90, also:-

x - y = 72 (given)

adding the 2 equations:-

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x = 81

and therefore

y + 81 = 90

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Answer the 2 angles are 81 and 9 degrees

5 0
3 years ago
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