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3 years ago

Find 3 ratios that are equivalent to the given ratio. 12 to 30

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

4 0

Answer:

4:10

24:60

48:120

As long as you multiply or divide both sides of the ratio by the same number, you will have an equivalent ratio.

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How to solve 2/3(6x-9)=-34

castortr0y [4]

$\frac{2}{3} (6x-9)=-34 \\ \\ \frac{2(6x - 9)}{3} =-34 \\ \\ \frac{2 \times 3(2x-3)}{3} =-34 \\ \\ \frac{6(2x - 3)}{3} = -34 \\ \\ 2(2x - 3) = -34 \\ \\ 2x - 3 - \frac{34}{2} \\ \\ 2x - 3 = -17 \\ \\ 2x -17 + 3 \\ \\ 2x = -14 \\ \\ x = - \frac{14}{2} \\ \\ x = -7 \\ \\ Answer: \fbox {x = -7}$

5 0

3 years ago

Solve the equation X/6 +4=16

elena55 [62]

Answer:

$x=160$

Step-by-step explanation:

$\frac{x}{6+4} =16$

$\frac{x}{10} =16$

$x=16 \times 10$

$x=160$

4 0

3 years ago

Read 2 more answers

For the function g(x)=3-8(1/4)^2-x

Reika [66]

Using function concepts, it is found that:

a) The y-intercept is y = 2.5.

b) The horizontal asymptote is x = 3.

c) The function is decreasing.

d) The domain is $(-\infty,\infty)$ and the range is $(-\infty,3)$ .
e) The graph is given at the end of the answer.

------------------------------------

The given function is:

$g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}$

------------------------------------

Question a:

The y-intercept is g(0), thus:

$g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5$

The y-intercept is y = 2.5.

------------------------------------

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3$

--------------------------------------------------

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty$

Thus, the horizontal asymptote is x = 3.

--------------------------------------------------

Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

--------------------------------------------------

Question d:

Exponential function has no restrictions in the domain, so it is all real values, that is $(-\infty,\infty)$ .
From the limits in item c, the range is: $(-\infty,3)$

--------------------------------------------------

The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

8 0

3 years ago

I need help with d 1 to d 3. I don't understand what to do.

erik [133]

What????

You just add 1 +3 =4

3 0

3 years ago

Please help

Gnesinka [82]

Answer:

1. subtract 9 from z- z-9

2. add 9 and z - z+9

3. divide 9 by z - 9/z

4. multiply 9 by z - 9z

Step-by-step explanation:

Your welcome!

Brainliest Appreciated!

7 0

3 years ago

Read 2 more answers