Ask question

Ask question

All categories

2 years ago

15

Solve. 10x^2 - 6 = 9x

1 answer:

grin007 [14]2 years ago

7 0

Answer:

B

Step-by-step explanation:

use quadratic formula: (-b ±√b²-4ac) ÷ 2a

a = 10, b = -9, c = -6

You might be interested in

What is the sum of x^2 - 3x + 7 and 3x^2 + 5x - 9

Dmitrij [34]

Hey there!

The answer to your question is $4x^2+2x-2$

Given:

$x^2-3x+7$ $and$ $3x^2 +5x-9$

First, we can start by adding like terms. The following are like terms:

$x^2$ $and$ $3x^2$

$-3x$ $and$ $5x$

$7$ $and$ $-9$

Now, we can add these pairs together, giving us:

$4x^2$

$2x$

$-2$

Now, add all this together, and we get our answer as:

$4x^2 + 2x - 2$

Have a terrificly amazing day!

7 0

2 years ago

PLSSSS HELPPPPPPPPPPPPPPPPPP

MArishka [77]

Answer:

$\huge\boxed{\sf Option \ 1}$

Step-by-step explanation:

People surveyed = 42

People who approved = 25

<u>So, the sample is:</u>

25 residents who approve.

<u>And population:</u>

42 residents surveyed.

$\rule[225]{225}{2}$

3 0

2 years ago

What is 2n-4>6 in the inequality term?

Kobotan [32]

2n - 4 > 6
    + 4 + 4
   2n > 10
      2 2
      n > 5

4 0

2 years ago

Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

<u>Left part:</u>

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

<u>Right part:</u>

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

3 0

3 years ago

What is the value of the expression

Stels [109]

$(-\dfrac{3^2}{3^3} )^2 = (-\dfrac{1}{3} )^2 = \dfrac{1}{9}$

OR (in more details):

$(-\dfrac{3^2}{3^3} )^2 = (-3^{2-3})^2 = (-3^{-1})^2 = (-\dfrac{1}{3} )^2 = \dfrac{1}{9}$

5 0

3 years ago

Read 2 more answers