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algol13
3 years ago
9

48-5+2x3+6.

Mathematics
1 answer:
kkurt [141]3 years ago
3 0
The parentheses must be placed here 48 - (5 + 2) * 3 + 6 in order for the equation to equal 33.
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How do you integrate arctan(x dx? i think that if you simplify the integral you get:?
Temka [501]
Integration by parts will help here. Letting u=\arctan x and \mathrm dv=\mathrm dx, you end up with \mathrm du=\dfrac{\mathrm dx}{1+x^2} and v=x. Now

\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du
\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx

For the remaining integral, setting y=1+x^2 gives \dfrac{\mathrm dy}2=x\,\mathrm dx, so

\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C

Putting everything together, you end up with

\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C
6 0
3 years ago
If you know the diameter of a circle, which statement describes how to find the area?
Rudik [331]
Well to find the area of a circle Pie times r^2. But If you already have the diameter, you could either split the diameter in half, and do the regulaur formula, OR you could ude this formula ----> Pie times The diameter
7 0
2 years ago
Read 2 more answers
SOMEONE PLZZ HELP
saul85 [17]
Square root of 27.  

3^(3/2) is the same as square root of 3 cubed, which is 27.  so square root of 27.

6 0
3 years ago
A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 50 percent of its previous height. The total vert
xz_007 [3.2K]
After every drop,the ball bounces to half it's previous height. With that understood.

1st drop -The ball drops 10m

1st bounce - 5m up
2nd drop - 5m down

2nd bounce - 2.5m up
3rd drop - 2.5m down

3rd bounce - 1.25m up
4th drop - 1.25m down

4th bounce - 0.625m up
5th/last drop - 0.625m down

To find the total vertical distance, you add them all.

10+5+5+2.5+2.5+1.25+1.25+0.625+0.625
=29.25m travelled in all.
6 0
3 years ago
Which best describes the graphs of the line that passes through (−12, 15) and (4, −5), and the line that passes through (−8, −9)
konstantin123 [22]

Answer:

C) They are perpendicular lines.

Step-by-step explanation:

We first need to find the slope of the graph of the lines passing through these points using:

m =  \frac{y_2-y_1}{x_2-x_1}

The slope of the line that passes through (−12, 15) and (4, −5) is

m_{1} =  \frac{ - 5 - 15}{4 -  - 12}

m_{1} =  \frac{ - 20}{16}  =  -  \frac{5}{4}

The slope of the line going through (−8, −9) and (16, 21) is

m_{2} =  \frac{21 -  - 9}{16 -  - 8}

m_{2} =  \frac{21  + 9}{16  + 8}

m_{2} =  \frac{30}{24}  =  \frac{5}{4}

The product of the two slopes is

m_{1} \times m_{2} =  -  \frac{4}{5}  \times  \frac{5}{4}  =  - 1

Since

m_{1} \times m_{2} =  - 1

the two lines are perpendicular.

4 0
3 years ago
Read 2 more answers
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