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evablogger [386]
3 years ago
12

What is the total area of William’s backyard in square feet?

Mathematics
2 answers:
professor190 [17]3 years ago
5 0

Answer:why

Step-by-step explanation:

because i don't know

Anton [14]3 years ago
5 0
Where’s the picture?
You might be interested in
Pls help I’m having trouble
kenny6666 [7]

Answer:

1161 ft

Step-by-step explanation:

36 ft with the triangles, 900 ft with the squares, 225 ft for the bottom square, so total is 1161 ft.

3 0
3 years ago
HELP! ASAP. 12 POINTS.
Alinara [238K]
I think the answer is c

7 0
3 years ago
Solve only if you know the solution and show work.
SashulF [63]
\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx

For the remaining integral, let t=\tan\dfrac x2. Then

\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}
\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}

and

\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt

Now the integral is

\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt

The first integral is trivial, so we'll focus on the latter one. You have

\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt

Decompose the integrand into partial fractions:

\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}

so you have

\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt

which are all standard integrals. You end up with

\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt
=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C
=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C
=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C

To try to get the terms to match up with the available answers, let's add and subtract \ln\left|1+\tan\dfrac x2\right| to get

2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C
2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C

which suggests A may be the answer. To make sure this is the case, show that

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1

You have

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}

So in the corresponding term of the antiderivative, you get

\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|
=\ln2-\ln|\cos x+\sin x+1|

The \ln2 term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C
5 0
3 years ago
Given the following sequence, find the 23rd term: 10.5, 11, 11.5, 12, 12.5, . . .
4vir4ik [10]
10.5, 11, 11.5, 12, 12.5...this is an arithmetic sequence with a common difference of 0.5

an = a1 + (n - 1) * d
n = term to find = 23
a1 = first term = 10.5
d = common difference = 0.5

sub and solve

a(23) = 10.5 + (23 - 1) * 0.5
a(23) = 10.5 + 22 * 0.5
a(23) = 10.5 + 11
a(23) = 21.5 <===
6 0
3 years ago
Angelo and Deborah are landscapers. Angelo charges a one time fee of $50 plus $15 per hour. Deborah charges a one time fee of $2
Tems11 [23]

Answer:

5 hours

Step-by-step explanation:

Step one:'

given data

Angelo charges a one time fee of $50 plus $15 per hour

let the total charge be y

and the number of hours be x

y= 50+15x----------------1

Deborah charges a one time fee of $25 plus $20 per hour

y= 25+20x----------------2

equate 1 and 2 to find x

50+15x=25+20x

collect like terms

50-25= 20x-15x

25= 5x

divide both sides by 5

x= 25/5

x= 5 hours

4 0
3 years ago
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