Please solve this quickly !

Number 6,8, and 10.

Aliun [14]

I think number 10 is 4/9

6 0

3 years ago

What is the simplified expression for,

Nostrana [21]

Here are a few rules you need to know for this equation:

Multiplying exponents of the same base: $x^m\times x^n=x^{m+n}$
Dividing exponents of the same base: $\frac{x^m}{x^n}=x^{m-n}$
Turning a negative exponent to a positive one: $x^{-m}=\frac{1}{x^m};\frac{1}{x^{-m}}=x^m$

So this is our algebraic expression: $\frac{3^{-4}\times 2^3\times 3^2}{2^4\times 3^{-3}}$

Firstly, multiply 3^-4 and 3^2: $\frac{3^{-4+2}\times 2^3}{2^4\times 3^{-3}}=\frac{3^{-2}\times 2^3}{2^4\times 3^{-3}}$

Next, divide:

$\frac{3^{-2}\times 2^3}{2^4\times 3^{-3}}=3^{-2-(-3)}2^{3-4}=3^12^{-1}$

Next, turn the negative exponent into a positive one: $3^12^{-1}= \frac{3}{2}$

<u>Your final answer is 3/2, or 1.5.</u>

6 0

3 years ago

It is estimated that 26% of the population of Gingerland have red hair. If 10 residents are chosen at random, what is the probab

sveticcg [70]

Answer:

The answer for me would be a) exactly 4 of them have red hair

Step-by-step explanation:

being that there is 26 percent of people in Gingerland that gave red hair I cam assume that that is a lot of people being that Gingelrand is like a city/town so exactly 4 of them is what I assume is the amswer

6 0

3 years ago

Many televisions are classified by the length of the diagonal of their screens. A 60 in. television has a screen height of 32 in

Colt1911 [192]

The 60 inch is diagonal of or the hypotenuse of the TV so we are going to use a^2+b^2=c^2 and let width to be a, then we solve the problem.
a^2=c^2-b^2
oh, take the square root of both sides.
a=((c^2-b^2))^(1/2)
plug 60 for c, 32 for b, and the Width is equal to 50.75 for significant figures.

7 0

3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

3 0

3 years ago