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3 years ago

Solve each system by the addition method.

Mathematics

1 answer:

bagirrra123 [75]3 years ago

6 0

Answer:

(x, y) = (2, -1.5)

Step-by-step explanation:

x + 2y = -1

5x - 4y = 16

<=>

(2)x + (2)2y = (2)(-1)

5x - 4y = 16

<=>

2x + 4y = -2

5x - 4y = 16

<=>

7x = 14

<=>

x = 2

(*)x + 2y = -1

=> 2 + 2y = -1

=> 2y = -3

=> y = -1.5

=> (x, y) = (2, -1.5)

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How can you add 3over 10 and 2over5

otez555 [7]

Answer:

7/10

Step-by-step explanation:

3/10+2/5

You have to make like denominators

So you multiply 2/5 by 2 since 1/ is the common denominator.

Then you add them

Hope this helped

6 0

3 years ago

Read 2 more answers

Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0

Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

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

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

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

$h(x)=\dfrac{1}{3}|x-6|+4$

Put the values of x to the equation of the function h(x):

$x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6$