That question is impossible because we don’t know what we are trying to solve for it has to be x or y in this scenario it can’t be both we need one to find the other

4 0

3 years ago

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I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and pr

Illusion [34]

Answer:

<h2><u>question 1</u></h2>

$n^{2} - 20n -96 = 0$

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence <u>n = 24 and n = -4 </u>

<u></u>

<h2><u>Question 2 </u></h2>

in the form $ax^{2} +bx +c = 0$

= $x^{2} +12x - 48$

make use of the formula :

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

replace values to make 2 equations :

1. $\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}$ = 3.17

2. $\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}$ = -15.2

hence <u>x = 3.17 and x = -15.2</u>

<u />

<h2><u>Question 3 </u></h2>

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence<u> x = 10 and x = 4</u>

<u />

<h2><u>Question 4 </u></h2>

in the form $ax^{2} +bx +c = 0$

this becomes $5b^{2} -20b-18-7$

= $5b^{2} -20b-25$

can simplify by 5

= $b^{2} -4b-5 =0\\$

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence <u>b = 5 and b = -1</u>

5 0

4 years ago

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