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AleksandrR [38]
2 years ago
8

Help Help Help Help Help FAST GO ZOOOOOOOOOOOOM WITH THIS QUESTION PLEASE

Mathematics
2 answers:
JulsSmile [24]2 years ago
5 0
60+x=90 all complementary angles add up to 90 degrees
xxMikexx [17]2 years ago
3 0

Answer:

A. 60+x=90

Step-by-step explanation:

Complimentary angles are angles that add to 90.

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Janie receives an allowance of $ 3 $3dollar sign, 3 per week. In addition, she can earn $ 2 $2dollar sign, 2 for each chore she
Gnesinka [82]

Hello  Dafnemlp47vj7

The inequality for one week would be <u><em>13  = 2c + 3 </em></u>

6 0
3 years ago
Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,
trapecia [35]
Not much can be done without knowing what \mathbf F(x,y,z) is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)
C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)

where 0\le t_1\le\dfrac\pi2 and 0\le t_2\le1. You have

\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2

and so the work is given by the integral

\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r
=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
{}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2
5 0
3 years ago
KL = 8x-5 and LM = 6x+3
Burka [1]
The answer is x=4. 
set the equations up: 8x-5=6x-3
Subtract 6x from one side and then the other side 2x-5=3 
Add 5 from one side and then the other 2x=8
Divide 2 on each side which gets you x=4
4 0
3 years ago
Read 2 more answers
hmu if ur bored and wana talk. age limit tho. cant be younger than 14 or older than 19 dont wanna talk to creepy people lol
Orlov [11]

Answfunny lol lol

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

and the integral of \frac1{x^2} converges over the same domain, so this integral must also converge by comparison.

We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

8 0
3 years ago
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