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3 years ago

6 & 7 please help me show work please and thank you!

Mathematics

1 answer:

lara31 [8.8K]3 years ago

3 0

Answer:

6. $\frac{29}{15}$ or 1 $\frac{14}{15}$

7. $\frac{133}{96}$ or 1 $\frac{37}{96}$

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please help! functions and relations. f(x)= square root of x-4. find the inverse of f(x) and it’s domain.

likoan [24]

ANSWER:

$\:D.\text{ }f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge-4$

STEP-BY-STEP EXPLANATION:

We have the following equation:

$f(x)=\sqrt{x}-4$

The inverse is the following (we calculate it by replacing f(x) by x and x by f(x)):

$\begin{gathered} x=\sqrt{f^{-1}(x)}-4 \\ \\ \sqrt{f^{-1}(x)}=x+4 \\ \\ f^{-1}(x)=(x+4)^2 \end{gathered}$

The domain would be the range of the original equation, and it would be the range of values that f(x) could take, which was from -4 to positive infinity, that is, f(x) ≥ -4.

Therefore, the domain is x ≥ -4.

So the correct answer is D.

$\:f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge -4$

4 0

1 year ago

I don't need you to work it out, just theorem(s) i need to reach the answer :)

VikaD [51]

Not sure why such an old question is showing up on my feed...

Anyway, let $x=\tan^{-1}\dfrac43$ and $y=\sin^{-1}\dfrac35$ . Then we want to find the exact value of $\cos(x-y)$ .

Use the angle difference identity:

$\cos(x-y)=\cos x\cos y+\sin x\sin y$

and right away we find $\sin y=\dfrac35$ . By the Pythagorean theorem, we also find $\cos y=\dfrac45$ . (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if $\tan x=\dfrac43$ , then (by Pythagorean theorem) $\sec x=\dfrac53$ , so $\cos x=\dfrac35$ . And from this, $\sin x=\dfrac45$ .

So,

$\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}$

7 0

4 years ago

X/-6= -20. What is x?

lesya692 [45]

The answer is x = 120

6 0

3 years ago

Read 2 more answers

Which equation represents y = −x2 + 6x + 7 in vertex form? y = −(x + 3)2 + 4 y = −(x − 3)2 + 10 y = −(x + 3)2 + 2 y = −(x − 3)2

Colt1911 [192]

$\text{The vertex form:}\ y=a(x-h)^2+k\\\\f(x)=ax^2+bx+c\to h=\dfrac{-b}{2a},\ k=f(h)\\\\\text{We have}\ y=-x^2+6x+7\to f(x)=-x^2+6x+7\\\\a=-1,\ b=6,\ c=7\\\\h=\dfrac{-6}{2(-1)}=\dfrac{-6}{-2}=3\\\\k=f(3)=-3^2+6(3)+7=-9+18+7=16\\\\Answer:\ \boxed{y=-(x-3)^2+16}$

8 0

3 years ago

[3 - (4 + 32 * 8) ÷ 4] =

serg [7]

<span>[3 - (4 + 32 * 8) ÷ 4]
=</span>[3 - (4 + 4) ÷ 4]
= 3 - 8 ÷ 4
= 3 - 2
= 1

7 0

3 years ago