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4 years ago

The lengths of the sides of a triangle are in the ratio

1 answer:

5 0

HOW TO SOLVE:

1) use x as a variable
2) the starting equation is:
  4x + 3x + 5x = 18

3) add all the like terms on the left side of the equation:
  12x = 18

4) Divide by 12 on both sides:
  x=1.5

5) Find the lengths of the sides:

- 4x = 4(1.5) = 6

- 3x = 3(1.5) = 4.5

- 5x = 5(1.5) = 7.5

SO, the lengths of the sides of this triangle are: 6, 4.5, and 7.5

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A fitness center charges $3 per visit plus a monthly fee of $30. How much would it cost in one month if Ella has 18 visits?

suter [353]

Her total should be 84. You have your 30 dollars that will not change which you need to add to the amount of every visit. To find the amount for each time she went you have to do 3*18 because she went 18 times for 3 dollars each. Your answer to 3*18 should be 54. You now have to add those 54 to the 30 per month. 30+54. That should be 84

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3 years ago

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Answer:

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2 years ago

1. Define sets

Rudik [331]

Answer:

Set is a collection of specified object.
The types of set are Finite Set. A set which contains a definite number of elements is called a finite set. ...

Infinite Set. A set which contains infinite number of elements is called an infinite set. ...

Subset. ...

Proper Subset. ...

Universal Set. ...

Empty Set or Null Set. ...

Singleton Set or Unit Set. ...

Equal Set.

The operation used on sets are. 1 Intersection 2.Union 3.Difference 4.Compelement

4 0

3 years ago

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

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

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

3 years ago

ect the best answer for the question. 3. Rewrite the equation x + y + 3 = –2x + 4y + 18 into standard form (ax + by = c). A. 3x

trapecia [35]

Answer:

Step-by-step explanation:

To rewrite the equation x + y + 3 = -2x + 4y + 18, first combine like terms.

x + y + 3 = -2x + 4y + 18

3x - 3y + 3 = 18

3x - 3y = 15

This is the standard form.

It can also be simplified by dividing each term by 3 to x - y = 5.

5 0

3 years ago