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3 years ago

I need help factoring this expression

Mathematics

1 answer:

omeli [17]3 years ago

3 0

X^2(x+2)-4(x+2)

(x^2-4) (x+2) = 0

do you have to keep going/?

or thats all they want?

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5x + 9 equals 2x + 12. What is the value of x?

jenyasd209 [6]

Answer: x = 1

Step-by-step explanation: When we have this kind of setup, we want to put our variables together on one side of the equation and our numbers together on the other side of the equation.

So first, let's put our variables together on the left side

by subtracting 2x from both sides of the equation.

That gives us 3x + 9 = 12.

Now we can move our numbers to the right by

subtracting 9 from both sides and we get 3x = 3.

Now divide both sides by 3 and x = 1.

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3 years ago

X-intercept of 5 and a y-intercept of -4. Write this street's equation in slope intercept form.

cricket20 [7]

Answer:

y = 5x -4 or you could also write y = 5x + -4

it's still the same thing.

Step-by-step explanation:

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3 years ago

James usually drinks 38 ounces of water per day. He read that he should drink 64 ounces of water per day. If he starts drinking

vova2212 [387]

Answer at very bottom of picture above

if explanation need comment

7 0

3 years ago

Using the attached link below.

Serhud [2]

$\text{Use the Pythagorean theorem:}\\\\r-hypotenuse\\\\r^2=7^2+5^2\\\\r^2=49+25\\\\r^2=74\to r=\sqrt{74}$

$\sin=\dfrac{opposite}{hypotenuse}\\\\\cos=\dfrac{adjacent}{hypotenuse}\\\\\tan=\dfrac{opposite}{adjacent}\\\\\cot=\dfrac{adjacent}{opposite}\\\\\text{We have}\\\\for\ the\ angle\ y:\\\text{opposite = 7}\\\text{adjacent = 5}\\\text{hypotenuse = }\ \sqrt{74}\\\\for\ the\ angle\ x:\\\text{opposite = 5}\\\text{adjacent = 7}\\\text{hypotenuse = }\ \sqrt{74}$

$\csc x=\dfrac{1}{\sin x}\\\\\sec x=\dfrac{1}{\cos x}$

$A.\\\\\sin x=\dfrac{5}{\sqrt{74}}=\dfrac{5\sqrt{74}}{74}\\\\\csc y=\dfrac{1}{\frac{7}{\sqrt{74}}}=\dfrac{\sqrt{74}}{7}\\\\B.\\\\\tan x=\dfrac{5}{7}\\\\\cot y=\dfrc{5}{7}\\\\C.\\\\\cos x=\dfrac{7}{\sqrt{74}}=\dfrac{7\sqrt{74}}{7}\\\\\sec y=\dfrac{1}{\frac{5}{\sqrt{74}}}=\dfrac{\sqrt{74}}{5}$

$D.\\\sin\theta=\dfrac{2}{3}\\\\\sin^2\theta+\cos^2\theta=1\to\left(\dfrac{2}{3}\right)^2+\cos^2\theta=1\\\\\dfrac{4}{9}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{4}{9}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{5}{9}\to\cos\theta=\sqrt{\dfrac{5}{9}}\to\cos\theta=\dfrac{\sqrt5}{3}\\\\\tan\theta=\dfrac{\sin\theta}{\cos\theta}\to\tan\theta=\dfrac{\frac{2}{3}}{\frac{\sqrt5}{3}}=\dfrac{2}{3}\cdot\dfrac{3}{\sqrt5}=\dfrac{2}{\sqrt5}=\dfrac{3\sqrt5}{5}$

6 0

3 years ago

-6/20 + 1.5 kdjbanskkcjdjxjxjdjhelp plz

Helga [31]

Answer:

1.2

Step-by-step explanation:

6 0

3 years ago