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expeople1 [14]
3 years ago
14

Ally makes $15 per hour. She works 40 hours a week. How much money will Ally make in February? Hint: February has 28 days.

Mathematics
2 answers:
Andreyy893 years ago
8 0
15x40=600
600x28=1,680
Angelina_Jolie [31]3 years ago
4 0
15x40= 600
600x4= 2,400 
i hope this is the correct answer. i multiplied 600x4 because she makes 600 dollars per week and there are 4 weeks in feburary
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Solve the polynomial: 2x³-11x²+18x-8=0
kotykmax [81]

Answer:

2/0.74

Step-by-step explanation:

To solve the polynomial, factorize it.

2x³-11x²+18x-8=0

After factorizing:

(x-2)(2x^2 - 7x +4) = 0

now, solve each bracket one by one:

x-2 = 0

x = 2

Then

2x^2 - 7x + 4 = 0

use quadratic formula t get 0.74

7 0
3 years ago
What is the answer to 18 - 3w = 4w
adelina 88 [10]

Answer:

Step-by-step explanation:

18 - 3w = 4w

Let's get all of the terms with the w on the right-hand side of the equation, and everything else on the left-hand side of the equation.

To do this, we should add 3w to both sides of the equation to remove the -3w from the left-hand side of the equation:

18 - 3w + 3w = 4w + 3w

18 = 7w

Finally, to get the w by itself, we can divide both sides by 7:

\frac{18}{7} = \frac{7w}{7}

w = \frac{18}{7}

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3 years ago
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Write -3/16, 7/4, -3/8 in order least to greatest
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3 years ago
Read 2 more answers
For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
3 years ago
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