All categories

2 years ago

Please help.....25 points!!!

Mathematics

2 answers:

ryzh [129]2 years ago

5 0

Answer:

Option 3

Step-by-step explanation:

y = ⅗x - 9

Make x the subject

⅗x = y + 9

x = 5y/3 + 15

Interswitch variables

g^-1(x) = 5x/3 + 15

kenny6666 [7]2 years ago

3 0

Answer:

Step-by-step explanation:

We want to find the inverse of g(x). One way to do this is to switch the g(x) and x in the original equation and then solve for g(x); that will give you the inverse.

Here, $g(x)=\frac{3}{5} x-9$ . Switch the g(x) and x to become: $x=\frac{3}{5} g(x)-9$ . Now solve for g(x) by isolating it:

$x=\frac{3}{5} g(x)-9$

$x+9=\frac{3}{5} g(x)$

$\frac{5}{3} * (x+9)= g(x)$

$g(x)=\frac{5}{3} x+15$

Thus, the answer is C.

Hope this helps!

You might be interested in

Parker rolled a number cube 50 times the three appeared 12 times what is the experimental probability of rolling a 3

horsena [70]

12/50 is the probability of rolling a 3
12/50= He has a 24% chance of rolling a three

3 0

3 years ago

68% of what number is 48

Gnesinka [82]

68*x/100=48
=> 68*x=4800
=> x=4800/68
=> x~=70.58

P.S: * means multiply, ~= means approximation, / means divide

8 0

3 years ago

If 15 x 332 = 4,980, then

DaniilM [7]

Answer:

then what?

Step-by-step explanation:

6 0

3 years ago

I need help! i’ll give brainliest

Yanka [14]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

If cos theta= -8/17 and theta is in quadrant 3, what is cos2 theta and tan2 theta

Karo-lina-s [1.5K]

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad 
\begin{array}{llll}
\textit{now, hypotenuse is always positive}\\
\textit{since it's just the radius}
\end{array}
\\\\\\
thus\qquad cos(\theta)=\cfrac{-8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}$

since the hypotenuse is just the radius unit, is never negative, so the - in front of 8/17 is likely the numerator's, or the adjacent's side

now, let us use the pythagorean theorem, to find the opposite side, or "b"

$\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}
\\\\\\
\pm\sqrt{17^2-(-8)^2}=b\implies \pm\sqrt{225}=b\implies \pm 15=b$

so... which is it then? +15 or -15? since the root gives us both, well
angle θ, we know is on the 3rd quadrant, on the 3rd quadrant, both, the adjacent(x) and the opposite(y) sides are negative, that means, -15 = b

so, now we know, a = -8, b = -15, and c = 17
let us plug those fellows in the double-angle identities then

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\\
2cos^2(\theta)-1
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\
cos(2\theta)=1-2sin^2(\theta)\implies cos(2\theta)=1-2\left( \cfrac{-15}{17} \right)^2
\\\\\\
cos(2\theta)=1-\cfrac{450}{289}\implies cos(2\theta)=-\cfrac{161}{289}$

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\left( \frac{-15}{-8} \right)}{1-\left( \frac{-15}{-8} \right)^2}
\\\\\\
tan(2\theta)=\cfrac{\frac{15}{4}}{1-\frac{225}{64}}\implies tan(2\theta)=\cfrac{\frac{15}{4}}{-\frac{161}{64}}
\\\\\\
tan(2\theta)=\cfrac{15}{4}\cdot \cfrac{-64}{161}\implies tan(2\theta)=-\cfrac{240}{161}$

6 0

3 years ago