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Elanso [62]
3 years ago
11

Given f(x) = 20x + 10. find f(5).Please as soon as possible​

Mathematics
1 answer:
Gemiola [76]3 years ago
6 0

Answer:

the answer is f(5) = 110..

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Can anyone help me with my math if not I'll just look it up on g o o g l e
Dahasolnce [82]

Answer:

Which math do you need to use help?

8 0
2 years ago
I need help on question 2 i need the simplified answer and the restrictions
Anastaziya [24]

Answer:

x cannot be 5 or -3.

\frac{x+5}{x+3}

Step-by-step explanation:

The restrictions for a fraction is that the bottom cannot be 0.

So if we find when the bottom is 0 we have found the values that x cannot be.

Let's solve x^2-2x-15=0.

Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.

Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.

So the factored form of the equation is:

(x-5)(x+3)=0

This means either x-5=0 or x+3=0.

We do have to solve both.

x-5=0 can be solved by adding 5 on both sides.

x-5+5=0+5

x+0=0+5

x=5

x+3=0 can be solved by subtracting 3 on both sides.

x+3-3=0-3

x+0=0-3

x=-3

So x can be any number except x=5 or x=-3.

We already factored the bottom as (x-5)(x+3).

The top is a difference of squares, x^2-a^2,

which can be factored as (x-a)(x+a).

So the top factors as (x-5)(x+5).

The fraction can then be written as:

\frac{x^2-25}{x^2-2x-15}=\frac{(x-5)(x+5)}{(x-5)(x+3)}

This can further simplified assuming x is not 5 we can write it as \frac{x+5}{x+3}.  I canceled the common factor of (x-5).

5 0
3 years ago
Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t
Anika [276]

Answer:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

Step-by-step explanation:

Given

\int\limits {\frac{x}{x^4 + 16}} \, dx

Required

Solve

Let

u = \frac{x^2}{4}

Differentiate

du = 2 * \frac{x^{2-1}}{4}\ dx

du = 2 * \frac{x}{4}\ dx

du = \frac{x}{2}\ dx

Make dx the subject

dx = \frac{2}{x}\ du

The given integral becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

Recall that: u = \frac{x^2}{4}

Make x^2 the subject

x^2= 4u

Square both sides

x^4= (4u)^2

x^4= 16u^2

Substitute 16u^2 for x^4 in \int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du

Simplify

\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

In standard integration

\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)

So, the expression becomes:

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)

Recall that: u = \frac{x^2}{4}

\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c

4 0
2 years ago
Jordan jarred 18 liters of jam after 6 days. How many days does Jordan need to spend making jam if he wants to jar 27 liters of
tiny-mole [99]

Answer:

69420

Step-by-step explanation:

69 420 equils 69 420

8 0
3 years ago
What is the factored equation for f(x)=x+x-6
tekilochka [14]

Answer:

f(x)=2(x-3) yup that's the answer

8 0
2 years ago
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