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kipiarov [429]
2 years ago
8

Can you chick are those right

Mathematics
1 answer:
Maksim231197 [3]2 years ago
4 0
I dont think we can see the whole problems
like half of it is cut off 

can you please fix it ?? 
thanks :3 
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Which tools can be used for inductive reasoning?
frosja888 [35]
Inductive reasoning is a kind of reasoning the involves taking specific information and from these information, will create a broader generalization or conclusion. This includes providing strong evidence for the truth of the conclusion. The tools that can be used for inductive reasoning are observation, experimentation and facts. 
6 0
3 years ago
Someone please help me with this problem
tamaranim1 [39]

Answer:

sorry my cumputer is not working

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A particle sits on a smooth surface and is acted upon by a time dependent horizontal force, giving it an
garik1379 [7]

(a) By the fundamental theorem of calculus,

<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>

The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives

<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²

(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of

∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024

7 0
3 years ago
If you are pushing on a box with a force of 20 N and there is a force of 7 N on the box due to sliding friction, what is the net
nika2105 [10]
13 N in the direction you are pushing it.

20 N------>
       <---7 N
=
13 N---->
3 0
3 years ago
Your friend says that the absolute value equation |3x+8|−9=−5 has no solution because the constant on the right side of the equa
xenn [34]

Solution, solve\:for\:x,\:\left|3x+8\right|-9=-5\quad :\quad x=-4\quad \mathrm{or}\quad \:x=-\frac{4}{3}

Steps:

\left|3x+8\right|-9=-5

\mathrm{Add\:}9\mathrm{\:to\:both\:sides},\\\left|3x+8\right|-9+9=-5+9

\mathrm{Simplify},\\\left|3x+8\right|=4

|f\left(x\right)|=a\quad \Rightarrow \:f\left(x\right)=-a\quad \mathrm{or}\quad \:f\left(x\right)=a,\\3x+8=-4\quad \quad \mathrm{or}\quad \:\quad \:3x+8=4

3x+8=-4,\\\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides},\\3x+8-8=-4-8,\\\mathrm{Simplify},\\3x=-12,\\\mathrm{Divide\:both\:sides\:by\:}3,\\\frac{3x}{3}=\frac{-12}{3},\\\mathrm{Simplify},\\x=-4

3x+8=4,\\\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides},\\3x+8-8=4-8,\\\mathrm{Simplify},\\3x=-4,\\\mathrm{Divide\:both\:sides\:by\:}3,\\\frac{3x}{3}=\frac{-4}{3},\\\mathrm{Simplify},\\x=-\frac{4}{3}

Combine\:the\:ranges,\\x=-4\quad \mathrm{or}\quad \:x=-\frac{4}{3}

\mathrm{The\:Correct\:Answer\:is\:No\:They\:Are\:Not\:Correct}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

4 0
2 years ago
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