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3 years ago

20 points if you help me on this proublem

Mathematics

1 answer:

VMariaS [17]3 years ago

3 0

Answer:

1) f(x) = x/8 - 7

2) f(x) = 1 - x/3

3) f(x) = 20 - 0.25x

4) f(x) = x + 7

Step-by-step explanation:

f^-1 = 8(x + 7)

x + 7 = y/8

x = y/8 - 7

f(x) = x/8 - 7

f^-1 = -3(x - 1)

y/-3 = x - 1

x = 1 - y/3

f(x) = 1 - x/3

f^-1 = 4(20 - x)

y/4 = 20 - x

x = 20 - y/4

f(x) = 20 - x/4

f(x) = 20 - 0.25x

f^-1 = x - 7

y = x - 7

x = y + 7

f(x) = x + 7

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4 years ago

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Kath starts with a number. she adds 3 to her number then double the result. finally, she subtracts 7. if she ended with 9, what

Dimas [21]

The answer is 5. Work backwards: first add 7 to 9, then divide by 2, then subtract 3.

9+7=16
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Cross Check:
5+3=8
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Final Answer: 5

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4 years ago

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with