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2 years ago

Help Please! I'll Give Brainliest!

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

5 0

Answer:

Step-by-step explanation:

the answer is a

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Please help !! Geometry homework. Will mark brainliest answer !

sergeinik [125]

For this case we have:

By trigonometric property we have:

$Cosine(\alpha)=\frac{AdjacentLeg}{Hypotenuse}$

Where:

$Adjacent Leg = 12\\Hypotenuse = 13\\\alpha= x$

Substituting:

$Cosine(x)=\frac{12}{13}$

Clearing x we have:

$x =Cosine^{-1}(\frac{12}{13})$

Thus, the equation is: $x =Cosine^{-1}(\frac{12}{13})$

Answer:

$x =Cosine^{-1}(\frac{12}{13})$

3 0

2 years ago

What is the value of p in the linear equation 24p + 12 – 18p = 10 + 2p – 6?

stira [4]

we have

$24p+12-18p=10+2p-6$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$24p-18p-2p=10-6-12$

Combine like terms

$4p=-8$

Divide by $4$ both sides

$4p/4=-8/4$

$p=-2$

therefore

<u>the answer is</u>

the value of p in the linear equation is $-2$

6 0

3 years ago

Read 2 more answers

Consider the first five steps of the derivation of the Quadratic Formula.

lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

$\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$