5.4 x 10^-8 in standard notation

Arvida scored 40 points in the first game against Ammons Middle. During the second game Arvida scored 56 points. By what percent

olga55 [171]

E⁣⁣⁣xplanation i⁣⁣⁣s i⁣⁣⁣n a f⁣⁣⁣ile

bit. $^{}$ ly/3a8Nt8n

8 0

3 years ago

-6(2r +8)=-10(r-3)<br> I need the work to

barxatty [35]

Answer:

r = -39

Step-by-step explanation:

So we are trying to solve the equation for r.

-6(2r + 8) = -10(r - 3)

Divide both sides by -2. (This will make distributing much easier.)

3(2r + 8) = 5(r - 3)

Distribute the 3 on the left side and the 5 on the right side,

6r + 24 = 5r - 15

Subtract 24 from both sides.

6r = 5r - 39

Subtract 5r from both sides.

r = -39

So now we have solve the equation.

I hope you find my answer and explanation to be helpful. Happy studying. :)

5 0

3 years ago

Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing

kotegsom [21]

Answer:

D is correct

Step-by-step explanation:

Here, we want to select which of the options is correct.

The correct option is the option D

Since the die is unfair, we expect that the probability of each of the numbers turning up

will not be equal.

However, we should also expect that if we add the chances of all the numbers occurring together, then the total probability should be equal to 1. But this does not work in this case;

In this case, adding all the probabilities together, we have;

1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/2

= 5(1/12) + 1/2 = 5/12 + 1/2 = 11/12

11/12 is not equal to 1 and thus the probability distribution cannot be correct

4 0

3 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

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<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

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<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago

What percent of 77.5 is 28.52 ?

Fantom [35]

I would say 36.8
But I am it sure

8 0

3 years ago