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3 years ago

Which of the following is the correct factorization of the trinomial below? -7x^2-5x+18

Mathematics

2 answers:

ryzh [129]3 years ago

8 0

Answer: C. (-7x+9)(x-2)

Step-by-step explanation:

1. Factor out the negative sign.

−(7x^2+5x−18)

2. Split the second term in 7x2+5x−18 into two terms.

−(7x^2+14x−9x−18)

3, Factor out common terms in the first two terms, then in the last two terms.

−(7x(x+2)−9(x+2))

4. Factor out the common term x+2x+2x+2.

−(x+2)(7x−9) or (-7x+9)(x-2)

DaniilM [7]3 years ago

7 0

Answer:

-1 (7x - 9) (x + 2)

Step-by-step explanation:

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Andy wants to buy a refrigerator for $769, a stove for $359, and a dishwasher for $309. round each cost to the nearest hundred t

Fiesta28 [93]

769 rounded to nearest 100 = 800

359 rounded to nearest 100 = 400

309 rounded to nearest 100 = 300

800 + 400 + 300 = $1,500 estimated total cost

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3 0

3 years ago

The height of a parallelogram is 5 feet more than it's base. If the area of the parallelogram is 204 square feet, what is its ba

siniylev [52]

The base is x.
The height is 5 feet more than the base, so it's x+5.
The area of a parallelogram is the product of its base and its height. The area is 204 square feet.

$x(x+5)=204 \\
x^2+5x=204 \\
x^2+5x-204=0 \\
x^2+17x-12x-204=0 \\
x(x+17)-12(x+17)=0 \\
(x-12)(x+17)=0 \\
x-12=0 \ \lor \ x+17=0 \\
x=12 \ \lor \ x=-17$

The length of the base of a parallelogram must be a positive number, so x=12.

$\hbox{the base:} \\ x=12 \\ \\
\hbox{the height:} \\ x+5=12+5=17$

The base is 12 feet long, the height is 17 feet long.

8 0

3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$