Solve by elimination 5x-15y=10 3x-5y=-6

1 answer:

6 0

Multiply bottom equation by -3 to add and eliminate y’s.

So...

5x-15y=10
+-9x+15y=18

You get -4x=28 so x=-7

Plug in to get y
5(-7)-15y=10
-35-15y=10
-15y=45
So y=-3

Check in other equation
3(-7)-5(-3)=
-21+15=-6

So x= -7 and y= -3

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A group of employees bought a get-well gift for a co-worker. The gift cost $16.06. If there are 22 employees in the group, how m

Kazeer [188]

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5 0

3 years ago

Read 2 more answers

2^5×8^4/16=2^5×(2^a)4/2^4=2^5×2^b/2^4=2^c A= B= C= Please I'm gonna fail math

aleksley [76]

9514 1404 393

Answer:

a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

(a^b)^c = a^(bc)

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You seem to have ...

$\dfrac{2^5\times8^4}{16}=\dfrac{2^5\times(2^3)^4}{2^4}\qquad (a=3)\\\\=\dfrac{2^5\times2^{3\cdot4}}{2^4}=\dfrac{2^5\times2^{12}}{2^4}\qquad (b=12)\\\\=2^{5+12-4}=2^{13}\qquad(c=13)$

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Additional comment

I find it easy to remember the rules of exponents by remembering that an exponent signifies repeated multiplication. It tells you how many times the base is a factor in the product.

$2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}$