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Elza [17]
3 years ago
8

Solve for c: b/c =a​

Mathematics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:

c = b/a

Step-by-step explanation:

b/c = a

b = ac

b/a = c

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Jamarius buys T-shirts for $6 each and marks up the price by 45%. How much profit does Jamarius make from each T-shirt sold?
3241004551 [841]

Answer:

$2.70

Step-by-step explanation:

To add 45 percent to the original price you do 100%+45%=145%=1.45

times 1.45  by the original price.

Then take away the original price from the answer you just got.

$1.45×$6=$8.70

$8.70-$6=$2.70

6 0
3 years ago
Solve the system by finding the determinants and using Cramer's Rule:<br> 2x - y = 4<br> 3x + y = 1
Ahat [919]

Answer:

<em>(1, - 2) </em>

Step-by-step explanation:

2x - y = 4

3x + y = 1

A = \left[\begin{array}{cc}2&-1\\3&1\end{array}\right] = 2(1) - 3( - 1) =2 + 3 = 5

A_{x} = \left[\begin{array}{cc}4&-1\\1&1\end{array}\right] = 4(1) - 1(- 1) = 4 + 1 = 5

A_{y} = \left[\begin{array}{cc}2&4\\3&1\end{array}\right] = 2(1) - 4(3) = 2 - 12 = - 10

<em>x </em>= \frac{A_{x} }{A} =<em> 1</em>

<em>y </em>= \frac{A_{y} }{A} = <em>- 2</em>

<em>(1, - 2)</em>

5 0
2 years ago
1. A bag contains the letters of the word “PROBABILITY”.
Mama L [17]

Since "PROBABILITY" has 11 letters in it, then every letter has 1/11 chances of getting picked, so the chances of getting an O tile would be 1/11, same for getting a B tile. But if you were to get both of them consecutively, then the chances would be 1/11 of 1/11 because you have 1/11 of a chance to get and O and then 1/11 of a chance to get a B. So you would be looking for 1/11 of 1/11, which means multiplying the divisor, 11*11=121. Therefore, you should have 1/121 chance of getting an O and B tile.

6 0
3 years ago
2. What is the approximate value of b in the triangle below?
s344n2d4d5 [400]

Answer:

d) 17.6

Step-by-step explanation:

Use the law of Sines

SinA/a=SinB/b=SinC/c

6 0
3 years ago
(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions
Sliva [168]
\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}
\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}
\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}
\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}
\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}
\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}
\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}
\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}
\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0
\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0

this will be continued
5 0
3 years ago
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