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Andrei [34K]
3 years ago
10

What are the factors of 40 and 55? *PLEASE SHOW WORK*

Mathematics
1 answer:
Vladimir [108]3 years ago
8 0

40- 1-40 , 2-20, 4-10,5-8

55- 1-55, 5,11

factors are numbers times each other put together to get the number

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Find the average value of the function f(t)=(t-2)^2 on [0,6]
pentagon [3]

Answer:

2

Step-by-step explanation:

The "average value of function f(x) on interval [a, b] is given by:

                      f(b) - f(a)

ave. value = ---------------

                         b - a

Here f(t)=(t-2)^2.

Thus, f(b) = (b - 2)^2.  For b = 6, we get:

          f(6) = 6^2 - 4(6) + 4, or f(6) = 36 - 24 + 4 = 16

For a = 0, we get:

           f(0) = (0 - 2)^2 = 4

Plugging these results into the ave. value function shown above, we get:

                      16 - 4

ave. value = ------------ = 12/6 = 2

                         6 - 0

The average value of the function f(t)=(t-2)^2 on [0,6] is 2.


4 0
3 years ago
Number 10 please please someone please help please please help
Tanzania [10]

Answer: A compound inequality contains at least two inequalities that are separated by either "and" or "or". The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities. A number is a solution to the compound inequality if the number is a solution to both inequalities.

3 0
3 years ago
NEED ASAP!!
iris [78.8K]

Hey!

------------------------------------------------

Answers:

B-B-A-B

A-B-B-B

------------------------------------------------

Explanation:

We assume that the winner is from a private school (B). As we look at the options we can see that there are two options that show B at least 3 times in the span.

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Hope This Helped! Good Luck!

6 0
3 years ago
Read 2 more answers
“encontrar la integral indefinida y verificar el resultado mediante derivación”
Oliga [24]

I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx

Haz la sustitución:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C

Para confirmar el resultado:

\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}

I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx

Sustituye:

y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx

\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C

(Te dejaré confirmar por ti mismo.)

I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx

Sustituye:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C

I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}

Sustituye:

u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}

\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C

Podemos hacer que esto se vea un poco mejor:

\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}

\implies I=-\dfrac{(t+1)^4}{4t^4}+C

4 0
3 years ago
Mr. Howard paid $99 for 18 feet of chain. At this rate,
vekshin1

Answer:

12.3 feet of chain is your answer.

3 0
3 years ago
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