Please help me i really need help please help me please i really need help please help me please

Mathematics

1 answer:

umka21 [38]3 years ago

6 0

Answer:

hello hello hello is there anybody in there

Step-by-step explanation:

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Write the equation in slope-intercept form. (-7,-2), (2,3)

Vadim26 [7]

Answer:

y = -5/9x + 4.1

(try doing this yourself first with different points to make sure this is the right answer, because im to lazy to check myself but ill try checking D:)

Step-by-step explanation:

first you have to find the slope . to do this use the equation y2 - y1/ x2 - x1

so write as -2 - 3/ -7 - 2 is equal to -5/-9

so now write y = -5/9x + b

we must now find the y-intercept so substitute one of the points given. lets substitute 2,3 it looks easier. So it will look like 3 = -5/9(2) + b

3 = -5/9(2) + b

3 = -10/9 + b

b = 3 + 10/9

b = 4.111111

um just round to the tenth place

6 0

2 years ago

Solve 2x^2 + x - 4 = 0 <br> X2 +

damaskus [11]

Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$