Mei has $100 in five-dollar and ten-dollar bills. She has a total of 11 bills. How many of each bill does she have?

Continue the pattern 12 24 72 288

riadik2000 [5.3K]

300 312 324 336 348 360

4 0

2 years ago

Please show me how you got the answer. Will give brainliest.

umka21 [38]

Answer: 410 possible points

To solve for the possible points in the course, use a proportion and cross multiply.

<h2>What is a proportion?</h2>

A proportion is two fractions that are equal to each other. For the fractions to be equal, they are proportionate. In our math problem, percentage is proportionate the the number of points.

The fractions represent information you have, and one variable will represent what you are solving for.

<h2 /><h3>Given</h3>

Your points = 361

Points to get an 'A' = your points + 8

'A' = 90% of possible points

<h3>Find the points needed for an 'A' at 90%</h3>

Points to get an 'A' = your points + 8

= 361 + 8

= 369

<h3>State variables</h3>

let <em>x</em> be the amount of possible points

<h3 /><h3>Write a proportion</h3>

We know the number of points for 90% is 369 points. Remember that 90% is the same as 90/100.

Keep the information with the same units inside the same fraction. Here, percentage is represented by the first fraction. Points is represented by the second fraction.

$\displaystyle{\frac{90}{100}=\frac{369}{x}}$

<h3>Solve</h3>

We can solve using cross multiplication. This is when we multiply each numerator by the denominator from the other fraction.

$90x = 369*100$ Simplify by multiplying.

$90x = 36,900$ Isolate the variable <em>x</em>.

$\displaystyle{\frac{90x}{90} = \frac{36,900}{90}}$ Divide both sides by 90 to cancel out 90 on the left.

$\displaystyle{x = \frac{36,900}{90}}$ Divide.

$x = 410$ Solved for <em>x</em>, representing the possible points.

∴ The possible points in the art course is 410.

Learn more about finding a missing number given percentage here: brainly.com/question/26535441

4 0

1 year ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .