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FromTheMoon [43]
3 years ago
13

What is the approximate area of a circle with a radius of 3ft? *

Mathematics
1 answer:
Ad libitum [116K]3 years ago
5 0
The answer is 28.26 because to find the area of a circle you need to do pie time radius squared.
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Simplify 9 exponent 2
STatiana [176]

Answer:

81

Step-by-step explanation:

9^2 = 81 bc 9x9=81

3 0
3 years ago
-5.8c+4.2-3.1+1.4c−5.8c+4.2−3.1+1.4c
Snowcat [4.5K]

Answer:

-10.2c + 2.2

Step-by-step explanation:

To make things less complicated, write down everything separate from the other. Then group together the intercepts and the coefficients into separate groups. After that, add them up separately in a format to where it will look like basic addition.

Hope this helps. I also apologize for any errors.

5 0
3 years ago
PLEASE HELP I DONT HAVE MUCH TIME LEFT!
DIA [1.3K]

Answer:

2.66

Step-by-step explanation:

5 0
2 years ago
Cos(x+y)/cos(x)sin(y)=cot(y)-tan(x)
zhuklara [117]
So there is an identity we'll need to use to solve this:

cos(x+y) = cosxcosy - sinxsiny

replace the numerator with the right hand side of that identity and we get:

(cosxcosy - sinxsiny)/cosxsiny

Separate the numerator into 2 fractions and we get:

cosxcosycosxsiny- sinxsiny/cosxsiny

the cosx's cancel on the left fraction, the siny's cancel on the right fraction and we're left with:

cosy/siny - sinx/cosx

which simplifies to:

coty - tanx




7 0
3 years ago
Find the average value of f over region
SVETLANKA909090 [29]
D is a right triangle with base length 1 and height 8, so the area of D is \dfrac12(1)(8)=4.

The average value of f(x,y) over D is given by the ratio

\dfrac{\displaystyle\iint_Df(x,y)\,\mathrm dA}{\displaystyle\iint_D\mathrm dA}

The denominator is just the area of D, which we already know. The average value is then simplified to

\displaystyle\frac74\iint_Dxy\,\mathrm dA

In the x,y-plane, we can describe the region D as all points (x,y) that lie between the lines y=0 and y=8x (the lines which coincide with the triangle's base and hypotenuse, respectively), taking 0\le x\le1. So, the integral is given by, and evaluates to,

\displaystyle\frac74\int_{x=0}^{x=1}\int_{y=0}^{y=8x}xy\,\mathrm dy\,\mathrm dx=\frac78\int_{x=0}^{x=1}xy^2\bigg|_{y=0}^{y=8x}\,\mathrm dx
=\displaystyle56\int_{x=0}^{x=1}x^3\,\mathrm dx
=14x^4\bigg|_{x=0}^{x=1}
=14
3 0
3 years ago
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