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3 years ago

9-x/5=3. Solve for x

Mathematics

1 answer:

kari74 [83]3 years ago

4 0

Answer:

$\boxed{\boxed{\sf x=-6}}$

Step-by-step explanation:

$\sf \cfrac{9-x}{5}=3$

$\sf \cfrac{5\left(9-x\right)}{5}=3\times \:5$

$\sf 9-x=15$

$\sf 9-x-9=15-9$

$\sf -x=6$

$\sf \cfrac{-x}{-1}=\cfrac{6}{-1}$

$\sf x=-6$

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Mariulka [41]

Answer:

1,305 remainder 0

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3 years ago

Ricardo earned $52,000 last year. He just received a 3% salary increase. What is his new yearly salary?

Juli2301 [7.4K]

<h3>Answer: 53,560 dollars</h3>

=======================================================

Explanation:

Having the salary increase by 3% means we can use the multiplier 1.03

Think of it like 100% + 3% = 1 + 0.03 = 1.03

The new salary is 52,000*1.03 = 53,560 dollars

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

The slightly longer way to do this is to first figure out how much 3% of 52,000 is

3% of 52,000 = 0.03*52,000 = 1560

Ricardo earns $1560 more dollars on top of the salary of $52,000

So overall his new salary is $1,560 + $52,000 = $53,560

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

Personally, I prefer the first method because it's quicker and it allows for multiple percentage increases to chain together. It also could be used to chain in percentage decreases as well.

8 0

3 years ago

Convert the polar equation r = 2 secØ to a Cartesian equation.y = 2x = 2x^2 = 2

ss7ja [257]

Polar and cartesian equation<h2>Initial explanation</h2>

Let's analyze the relation between r and x and y:

We have that between the indicated value of r (of the polar coordinates) and x and y (of the cartesian coordinates) there is a relation because they form a triangle. If r changes, then the value of x and y will change.

<h2>STEP 1: given equation</h2>

Using the given equation

r = 2 secØ

we have that

$\begin{gathered} r=2secØ \\ \downarrow \\ \frac{r}{2}=secØ \end{gathered}$ <h2>STEP 2: secØ equation</h2>

Observing the image of the initial explanation we have a right triangle, we know that the equation of

secØ for any right triangle is given by:

$\sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}$

In this case,

hypotenuse = r

adjacent side = x

then,

$\begin{gathered} \sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{r}{x} \\ \sec Ø=\frac{r}{x} \end{gathered}$ <h2>STEP 3: comparison between given equation and secØ equation</h2>

Then, we have that: