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

Halp plsss :3 tank chu! Mark as Brainliest!

aleksandrvk [35]

Answer:

_____________________________

l l

l____________________________l

8 0

3 years ago

Read 2 more answers

How do i write 14 ten thousands and 12 thousands in standard 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- 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 x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

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

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -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 f(x) is differentiable everywhere on its domain, except at the point x = -1.

4 0

3 years ago