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3 years ago

How do I use this app

Mathematics

2 answers:

aev [14]3 years ago

7 0

Answer:

You can answer people's questions by clicking on their question. If there question is not clear you can ask them about it. If you have a question then you can click on the ask question button on the homepage. If your answer is not clear someone can report it and it will erase it but you still get the points. If you plagiarize or do something else like that you can get a warning if you get multiple warnings then your account will be erased if your question is not clear then someone can ask you about it and then you can either answer them back or ignore it. You want to get more points and you get points by answering questions. (Hope this helps)

Natalka [10]3 years ago

4 0

Answer:

Ask a question and fellow persons will help you .

Step-by-step explanation:

Just type your question in the blue box at the top of course.

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What is two plus two

Olenka [21]

Answer:

2+2=4

Step-by-step explanation:

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3 years ago

Read 2 more answers

for the given kite find the length of side LM write your answer in simplest radical form. show all work.

Gnesinka [82]

Answer:

13sqrt(2) units

Step-by-step explanation:

LM² = 7² + 17²

LM² = 338

LM = sqrt(338)

LM = 13sqrt(2)

sqrt: square root

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3 years ago

Find the value of 2 devide3times16

Nikolay [14]

Answer:

32/3

Step-by-step explanation:

6 0

2 years ago

now find the components nx and ny of n⃗ in the tilted coordinate system of part b. express your answer in terms of the length of

saw5 [17]

The components are Nx= Ncosθ and Ny= -Nsinθ

<h3>What is a Vector?</h3>

We know that the vector quantities are those quantities that have magnitude as well as direction.

Each vector quantity can be divided into two parts a horizontal and vertical component, the vertical component is known as the sine component while the horizontal component is known as the cosine component.

A vector component is the product of its length and the component angle.

Generally, F sinθ is the vertical component and F cosθ is the horizontal component,

Now, from the diagram the horizontal component of vector 'r' is

Nx= Ncosθ

and, the vertical component will be

Ny= -Nsinθ

this is in the opposite direction

Learn more about vectors here:

brainly.com/question/1600633

#SPJ4

7 0

2 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

1 year ago