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Goshia [24]
3 years ago
5

PLZZZ HELP

Mathematics
2 answers:
Inessa05 [86]3 years ago
7 0
Multiplicative identity because it tells you that any number times 1 is equal to that number
Nezavi [6.7K]3 years ago
3 0
Multiplicative identity, which states that anything multiplied by 1 is itself.
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What is a prime number
UkoKoshka [18]

Answer:

A prime number is any number that cant be divide by any number but 1 (Ex. 1 2 3 5; Non-Ex. 4 6 8)

5 0
3 years ago
Read 2 more answers
PETS Becky wants to buy some fish for her aquarium. She has $20 to spend and the fish cost $2.50 each. Write and solve an inequa
Vilka [71]
To start, you want to determine the inequality. You can do this by setting the number of buying n fish to be less than or equal to her current amount of money, being as it can't go over or she wouldn't be able to afford the fish. This inequality would be written so her amount of money (20) would be greater than or equal to the cost of n fish (2.5n), which would then look as such: 20≥2.5n. To solve this inequality, just solve for n by dividing both sides by 2.5, giving you 8≥n. This would mean that Becky can afford to buy up to 8 fish.
6 0
3 years ago
Need help please any body
mash [69]
She should by the one that is written as 375/1000 inches because if Trudy needs a nail that measures 0.375 she should get that one and also .375 and 375/1000 is really the same thing. hope it helps
4 0
3 years ago
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
Which of the following segments is a diameter of O? <br><br> A. UF <br> B. WM <br> C. WO <br> D. OF
pishuonlain [190]

Answer:

The answer to your question is letter A

Step-by-step explanation:

In the picture, we can observe that

UF is the diameter of the circle. Diameter is a line that reaches from one point on the edge of a circle, through its center, to a point on the opposite side.

WM is a secant or a chord

WO is the radius

OF is also a radius

4 0
3 years ago
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