I need help please ASAP

David earned a total of $378.75 in 15 hours. How much did David earn per hour? NO GUESSES NO LINKS NO BOTS

enot [183]

Answer: $25.25

Step-by-step explanation:

378.75/15 = 25.25

3 0

2 years ago

Read 2 more answers

Solve all of them please u get 30 points

nika2105 [10]

Step-by-step explanation:

from which text book please

8 0

2 years ago

Read 2 more answers

Choose two angles that are each separately alternate exterior angles with 412.

Alborosie

Answer:

∠2 and ∠5

Step-by-step explanation:

we know that

<u>Alternate Exterior Angles</u> are a pair of angles on the outer side of each of those two lines but on opposite sides of the transversal

In this problem

∠12 and ∠2 are alternate exterior angles

∠12 and ∠5 are alternate exterior angles

therefore

∠2 and ∠5 are each separately alternate exterior angles with ∠12

6 0

3 years ago

What does the equal

sergejj [24]

Plug it in your y= screen of your graphing calculator and hit 2nd and then graph. it will give u a table of values

7 0

3 years ago

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Now subtract b/(2a) on both sides:

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Combine the fractions (they have the same denominator):

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

6 0

3 years ago