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3 years ago

How is this a nonlinear?

Mathematics

2 answers:

OlgaM077 [116]3 years ago

7 0

It’s not a straight line it’s a v !!

valina [46]3 years ago

4 0

Answer:

this is because its not in a straight line that goes through the origin

Step-by-step explanation:

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6th grade math help me pleaseeeee

Ivenika [448]

Answer:

This is the answer of your question. ☺☺

Step-by-step explanation:

24 -37*2

24-74

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3 years ago

The work of a student to find the dimensions of a rectangle of area 8 + 16x and width 4 is shown below:

Leya [2.2K]

Answer:

Step 1: 8 + 16x Step 2: 4(4) + 4(12x) Step 3: 4(4 + 12x) Step 4: Dimensions of the rectangle are 4 and 4 + 12x In which step did the student first make an error and what is the correct step? Step 3; 4 + (4 + 12x) Step 3; 4 + (4 ⋅ 12x) Step 2; 4(2) + 4(4x) Step 2; 4(2) + 4x(2)

3 0

2 years ago

qaws [65]

Answer:

(-2, 0.5) first

Step-by-step explanation:

X (-4, 1) → X' (-4*0.5, 1*0.5)

X'(-2, .5)

5 0

3 years ago

Answers for all questions

abruzzese [7]

Answer:

3 and dived 30 multy is 182 is your answer

Step-by-step explanation:

1+1=2

8 0

3 years ago

PLEASE HELP! I'll give 5 out of 5 stars, give thanks, and give as many points as I can.

Liula [17]

Answer:

$\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}$

Step-by-step explanation:

$\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)$

$\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)$

7 0

3 years ago

Read 2 more answers

How is this a nonlinear?​

How is this a nonlinear?