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3 years ago

What is (f⋅g)(x)?

Mathematics

2 answers:

makvit [3.9K]3 years ago

7 0

Answer: x^7-9x^4+9x^3+81

Step-by-step explanation:

there is the answer i did the test! hope it helps !!!!!!!!!!!!!!!!!!!!!!

vichka [17]3 years ago

5 0

The dot products of the two functions will be given as follows:
f(x)=x^4-9
g(x)=x^3+9
thus
(f*g)(x)
=(x^4-9)(x^3+9)
=x^3(x^4-9)+9(x^4-9)
=x^7-9x^3+9x^4-81
hence the answer is:
(f*g)(x)=x^7-9x63+9x^4-81

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Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or e

netineya [11]

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

$F(x)=\frac{e^x-1-x}{x^2}$

When,

$x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281$

$x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885$

$x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091$

$x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438$

$x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)$

Correct upto six decimal places.

Now,

$\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}$ $(\frac{0}{0})$ form, applying L-Hospital rule that is differentiating numerator and denominator we get,

$\lim_{x\to 0}F(x)$

$=\lim_{x\to 0}\frac{e^x-1}{2x}$ $(\frac{0}{0})$ form.

$=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)$

Limit exist and is 0.5. That is according to (1) we can see as the value of x lesser than 1 and tending to near 0, value of the function decreases respectively. And from (2) it shows ultimately it decreases and reach at 0.5, consider as limit point of F(x).

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3 years ago

Help pls I don’t get it

Dmitriy789 [7]

68.40 divide 6 = 11.4

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4 years ago

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Step-by-step explanation: