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3 years ago

PLEASE HELP FIRST GETS BRAINLIEST

Mathematics

1 answer:

atroni [7]3 years ago

5 0

Your answer would be 24 marbles. This is because if you add up all the number of pulls, you get 20, which means we would need to multiply all of the results by 3 to see how many pulls you would get in a bag of 60, and so 8 x 3 = 24. I hope this helps,

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Look at the picture below

Vadim26 [7]

= 2x + 11

Steps

(8x+9) -(6x-2)

Remove ( a ) = a

8x + 9 - 6x + 2

Simplify

8x+ 9 - 6x + 2: 2x + 11

= 2x + 11 <----- your answer

Please tell me if this was right

3 0

3 years ago

Integrate the following problem:

vazorg [7]

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

$u=e^-^x$
$du=-e^-^x dx$

$dv=cos2x \ dx$
$v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$ .

Let's find u, du, dv, and v for this integral:

$u=-e^-^x$
$du=e^-^x dx$
$dv=\frac{sin2x}{2} dx$
$v=\frac{-cos2x}{4}$

Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$

Make the fractions have common denominators.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} - \frac{e^-^x \cdot cos2x}{4}$

Simplify this equation.

$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}$

Multiply the right side by the reciprocal of 5/4.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}$

Factor $e^-^x$ out of the numerator.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}$

Simplify this by using exponential properties.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}$

The final answer is $\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C$ .

7 0

3 years ago

Read 2 more answers

A number is greater than 50 and less than 85 the number has 13 and 3 as factors

alina1380 [7]

Answer:

The Answer is: 78.

Step-by-step explanation:

This one makes you think!

Let n = the number.

The number is greater than 50 and less than 85:

50 < n < 85

Some of the factors are 13 and 3.

So, multiply 13 * 3 = 39.

Multiply again by 2:

39 * 2 = 78.

78 can be factored as:

13 * 3 * 2 = 78, and it is greater than 50 and less than 85.

The number is 78.

Hope this helps! Have an Awesome Day!! :-)

8 0

4 years ago

10. Jenna flips two pennies 105 times. How

Studentka2010 [4]

Answer:

Step-by-step explanation:

When you flip two coins, you can have 4 different outcomes:

All outcomes are equally likely, but notice that if you do not keep track of which coin is which, you will get 1/4 of the flips to be both heads, 1/4 of the flips to be both tails, and 1/2 of the flips to be one heads and one tails.

p(both heads) = 1/4

With 105 flips, the expected number of both heads is

105 * 1/4 = 26.25

Answer: 26

7 0

3 years ago

What is the minimum value

IgorC [24]

It's the smallest value in this equation which should at x=-10, the equation is x=-b/(2a)

3 0

3 years ago