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VMariaS [17]
3 years ago
7

What is the distance from the origin of point A graphed on the complex plane below?

Mathematics
2 answers:
masha68 [24]3 years ago
7 0

The correct Answer is B

stellarik [79]3 years ago
6 0
Use pythagoras theorem. square root of 13 the answer is b. h^2=3^2+2^2=13
h=
\sqrt{13}

.

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How can you be sure that 0.1 ÷ 100 = 0.001?
KATRIN_1 [288]

Answer: When dividing by 100, move the decimal two places to the left

Step-by-step explanation: 0.1 / 100 = 0.001

There are 2 zeros in 100 so you move the decimal to the left 2 times when you are dividing. When you move the decimal 2 times in 0.1 you get 0.001

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anzhelika [568]
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Jerry starts a new career making $32,000 annually. Each year for a total of 30 years
LUCKY_DIMON [66]

Step-by-step explanation:

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Which product has 4 zeros after the digit 3?
sineoko [7]
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3 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
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