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Oksana_A [137]
3 years ago
6

Plz help me if I get this wrong Im going to fail the class I will give the brainlist

Mathematics
1 answer:
Shtirlitz [24]3 years ago
4 0

Answer:

108

Step-by-step explanation:

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Halp plsss :3 tank chu! Mark as Brainliest!
aleksandrvk [35]

Answer:

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8 0
3 years ago
Read 2 more answers
How do i write 14 ten thousands and 12 thousands in standard<br> form
Burka [1]
In standard form:
answer- 14,000 and 12,000

Example
←LEFT 12,000
                 ↑
   comma(THOUSANDS PLACE)

ten thosands is the second number after the comma(thousands) to the left
8 0
3 years ago
Which number gives you an irrational product when you multiply it by 2
soldi70 [24.7K]
<h2>Explanation:</h2><h2></h2>

An irrational number is a number that can't be written as a simple fraction while a rational number is a number that can be written as the ratio of two integers, that is, as a simple fraction. So in this case we have the number 2 which is ration, and we can multiply it by an irrational number n such that the product is an irrational number. So any irrational number will meet our requirement because the product of any rational number and an irrational number will lead to an irrational number. For instance:

\bullet \ 2\sqrt{2} \\ \\ 2 \ is \ rational \\ \\ \sqrt{2} \ is \ irrational \\ \\ 2\sqrt{2} \ is \ irrational \\ \\ \\ \bullet \ 2\pi \\ \\ 2 \ is \ rational \\ \\ \pi \ is \ irrational \\ \\ 2\pi \ is \ irrational

6 0
3 years ago
Help pls- <br><br> I dont know it T^T
docker41 [41]

Answer:

Ans: H and A

Step-by-step explanation:

By the identity:

1-(cos(x))cos^{2}(x)+ sin^{2}(x)=1\\cos^{2}(x)=1- sin^{2}(x)\\\\cos(x)=\sqrt{1- sin^{2}(x)} \\sin(x)=\sqrt{1- cos^{2}(x)}\\\\\sqrt{1- cos^{2}(x)}/sinx=1\\\sqrt{1- sin^{2}(x)}/cosx=1\\\sqrt{1- cos^{2}(x)}/sinx+\sqrt{1- sin^{2}(x)}/cosx=2\\\\\\

Ans: H

For second quesiton, the transformation of a graph is that:

f(x) + k means vertical translation up by k units

Two curve have the same maximum value, so the graph doesn't translate upwards or downwards, thus b=0

Ans: A

7 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
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