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2 years ago

What is the factored form of this polynomial?

Mathematics

2 answers:

KatRina [158]2 years ago

7 0

The right answer is (x-4)(x-9)

look at the attached picture

Hope it helps...

Good luck on your assignment

oksian1 [2.3K]2 years ago

4 0

Answer:

$(x - 4)(x + 9)$

Answer C is correct.

Step-by-step explanation:

${x}^{2} + 5x - 36 \\ {x}^{2} + 9x - 4x - 36 \\ x(x + 9) - 4(x + 9) \\ = (x - 4)(x + 9)$

hope this helps

brainliest appreciated

good luck! have a nice day!

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Find an equation of variation in which y varies directly as x and y=420 when x= 100. Then find the value of y when x=11.

sweet [91]

Let's Simplify X to equal 1 by dividing by 100. then we divide 420 by 100 to get 4.2, so y=4.2 when x=1, we then multiply by each by 11.

y = 46.2
x = 11

4 0

3 years ago

Stephanie bought 36.5 inches of lumber at Home Depot the lumber cost 2.10 per inch including the sales tax. Stephanie bought 36.

tamaranim1 [39]

Answer:76.65

Step-by-step explanation:

First, you write 36.5 times 2.10

Then, multiply them.

That equals 76.65

4 0

3 years ago

Help with this please

olga2289 [7]

Answer:

( x + 2 ) and ( x - 11 )

Step-by-step explanation:

x² - 9x - 22 = 0

x² + 2x - 11x - 22 = 0

x ( x + 2 ) - 11 ( x + 2 ) = 0

( x + 2 ) ( x - 11 ) = 0

Factors of x² - 9x - 22 are ( x + 2 ) and ( x - 11 ).

8 0

3 years ago

Read 2 more answers

For sin2x+cosx=0, use a double-angle or half-angle formula to simplify the equation and then find all solutions of the equation

Archy [21]

Answer:

Step-by-step explanation:

Given is a trignometric equation in x, as

$sin2x+cosx=0$

TO make it in one trig ratio, we can replace sin2x as 2sinx cosx

WE get now

$2sinxcosx+cosx=0\\$

$cosx(2sinx+1)=0\\cosx=0, sinx =-0.5\\$

Principal solution is $x=\frac{\pi}{2} , \frac{-\pi}{6}$

x = ±π/2 + 2kπ, where k is any integer or

x=±pi/6 +k pi, where k is any integer.

General solution is

4 0

3 years ago

What is PA Enter your answer in the box

SSSSS [86.1K]

Answer:

Step-by-step explanation:

P is the in-center

⇒PA=PE=PD because they are in-radius of the in-circle

We know that, tangent segments drawn from a point outside the circle are always equal in length

⇒DK=EK=7.2

In right triangle PKE,

using Pythagoras' Theorem : $PK^{2}=PE^{2}+KE^{2}$

⇒ $PE^{2}=PK^{2}-KE^{2}$

⇒ $PE=\sqrt{PK^{2}-KE^{2} }$

⇒ $PE^{2}=\sqrt{7.8^{2}-7.2^{2}} }$

⇒ $PE=3$

Therefore, $PA=3$

3 0

3 years ago