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ololo11 [35]
3 years ago
6

Coordinates: A (1, 1) B (1, 3) C (4, 3), Rectangle ABCD has the coordinates shown. Find the coordinates of point D. A) (5,1) B)(

4, 2) C)(3, 1) D)(4, 1)
Mathematics
2 answers:
Allushta [10]3 years ago
8 0

Answer:

(4,1)

Step-by-step explanation:

Nataly [62]3 years ago
5 0
The answer to your question should be D
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∠1 and ∠2 are supplementary angles. If m∠1 = 77°, then m∠2 =
posledela

Answer:

B. 103°

Step-by-step explanation:

Supplementary angles add up to 180°, so we can find the measure of angle 2 by subtracting 77 from 180

180 - 77

= 103

So, m∠2 is 103°

7 0
3 years ago
Find the volume of the prism. 4 in. 8 in. 8 in. ​
Temka [501]

Answer:

256

Step-by-step explanation:

8×8=64

64×4=256

64

4

256

3 0
2 years ago
Read 2 more answers
What is the cheapest way to go to a ball game if the admissions options are as shown? A) $10 for a single B) $15 for a couple C)
vivado [14]

Answer: C

Step-by-step explanation: A = 10 for one person. B = 7.5 for one person. C = 6.67 for one person. D = 8.75 for one person. E = 10 for one person.

Out of all of these, C is the cheapest.

5 0
3 years ago
14 more than 6 times a number is<br> greater than or equal to 28
Studentka2010 [4]

Answer:

6x + 14 ≥ 28

Step-by-step explanation:

6 times a number (6x)

14 more than (so, add 14 to 6x)

greater than or equal to, so this symbol ≥

= 6x + 14 ≥ 28

SOLVED:

6x≥14

x≥2.333333333333333

4 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
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