3 years ago

Find an order pair (x,y) that is solution to the equation 3x-y=8

Mathematics

2 answers:

Ivahew [28]3 years ago

7 0

Answer:

(2.67,-8)

Step-by-step explanation:

You first make x 0 so it is -3x0 to cancel that out to find y. You then see that you get -y=8 so you make y positive by moving the negative to the other side to get y = -8. Then, you do the same type of process for x and get 2.67. I could be wrong but wanted help as much as I could!

bogdanovich [222]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Hazhair bought 2 DVDs for $12.95 each. He spent $25 on magazines. Did he spend more on DVDs or magazines? How much more? Write e

marshall27 [118]

Answer

5.9

12.95*2=25.9

25.9-20

3 0

3 years ago

Can someone answer these questions ? thank you

Naya [18.7K]

Answer:

Step-by-step explanation:

12a)To rationalize the denominator, multiply the denominator and numerator by √5.

$\frac{15}{\sqrt{5}}=\frac{15*\sqrt{5}}{\sqrt{5}*\sqrt{5}}\\\\=\frac{15\sqrt{5}}{5}\\\\=3\sqrt{5}$

b) (a+b)² = a² + 2ab + b²

(1 +√3)² = 1² + 2*1*√3 + (√3)²

= 1 + 2√3 + 3

= 4 + 2√3

a = 4 ; b =2

13) (a + b)(a - b) = a² - b²

$\frac{(6-\sqrt{5})(6+\sqrt{5})}{\sqrt{31}}=\frac{6^{2}-(\sqrt{5})^{2}}{\sqrt{31}}\\\\ =\frac{36-5}{\sqrt{31}}\\\\=\frac{31}{\sqrt{31}}\\\\=\frac{31*\sqrt{31}}{\sqrt{31}*\sqrt{31}}\\\\=\frac{31\sqrt{31}}{31}\\\\=\sqrt{31}$