What is the answer to this problem -3q+4=13

Where could point E be and D is between E and B?

Butoxors [25]

It is d.2. because its like a pattern you have to follow on the graph.

6 0

3 years ago

Define $g$ by $g(x)=5x-4$. If $g(x)=f^{-1}(x)-3$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $5a+5b$.

Slav-nsk [51]

Answer:

Step-by-step explanation:

Given g(x)=5x-4 and g(x)=f^{-1}(x)-3

Substituting g(x) = 5x-4 into the second equation we have;

5x-4 = f^{-1}(x)-3

f^{-1}(x) = 5x-4+3

f^{-1}(x) = 5x-1

To get f(x), let us first make y to be equal f^{-1}(x)

y = 5x-1

expressing x in terms of y to get f(x), we have;

5x = y+1

x = y/5+1/5

replacing y with x, we will have;

y = x/5 + 1/5

F(x) = x/5 + 1/5

Comparing x/5 + 1/5 with ax+b, a = 1/5 and b = 1/5

5a + 5b = 5(1/5)+ 5(1/5)

5a+5b = 1+1

5a+5b = 2

8 0

3 years ago

Solve the question and check the solution. Express numbers as integers or simplified fractions. 5(m+5)=7m+9

Yanka [14]

Answer:

m = 8

Step-by-step explanation:

5(m+5)=7m+9 (distribute & evaluate parenthesis)

5(m)+5(5)=7m+9

5m + 25 = 7m + 9 (move all m-terms to one side and constants to the other)

5m - 7m = 9 - 25 (evaluate both sides)

-2m = -16 (divide both sides by -2)

m = (-16) / ( -2) = 8

7 0

3 years ago

Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE

AVprozaik [17]

Answer:

D

Step-by-step explanation:

We are given that:

$\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi$

And we want to find the value of tan(2<em>x</em>).

Note that since <em>x</em> is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

Using double angle identities:

$\displaystyle \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}$

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(<em>x</em>) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

$\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since <em>x</em> is in QII, cosine and tangent will be negative while sine will be positive. Hence: