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yan [13]
2 years ago
8

Paisley is going to invest in an account paying an interest rate

Mathematics
1 answer:
PolarNik [594]2 years ago
3 0

Answer:

She needs to do 23 sex a day and big pssyyyyyy

Step-by-step explanation:

uhhhhh

You might be interested in
9. Solve for x: 2x – y = (3/4)x + 6.​
AlekseyPX

Answer:

x = 4(y + 6)/5

Step-by-step explanation:

2x - y = (3/4)x + 6

2x - (3/4)x = y + 6

(8x - 3x)/4 = y + 6

5x/4 = y + 6

5x = 4(y + 6)

5x = 4y + 24

x = (4y + 24)/5

Therefore, x = 4(y + 6)/5

4 0
3 years ago
Manipulate the triangle so angle A measures 41° and
Sergio039 [100]

Answer: angle B will be 90°

Sum of the angles = 180°

4 0
3 years ago
Read 2 more answers
What is the eighth term of the geometric sequence 6, 18, 54
Rainbow [258]

Answer:

= 4374.

Step-by-step explanation:

 it is important to understand the pattern hidden in such problem.Let’s give it a try 2, 6, 18, 54 and so on.It can be written as 2, 3*2, 9*2, 27*2 and so on.

This can be further written as2 (1, 3, 9, 27, and so on) as 2 is common in every term.Now if you see the chain 1,3,9, 27 and so….you will see a pattern hidden i.e. 3=1*3 ,9=1*3*3, 27=1*3*3*3 now 27 is the 4th term consist of three 3. So 8th term would consist of seven 3. 8th would be 8th term = 1*3*3*3*3*3*3*3 = 2187 Hence the 8th term for the series 2,6,18,54 would be= 2*2187 = 4374.

4 0
3 years ago
A class of 25 students took a spelling test. Two students scored 100 on each test, nine students scored 95 on each test, ten stu
lawyer [7]
The average score is 90.6
5 0
3 years ago
Read 2 more answers
Evaluate the line integral for x^2yds where c is the top hal fo the circle x62 _y^2 = 9
natulia [17]
Parameterize C by

\mathbf r(t)=\langle x(t),y(t)\rangle=\langle3\cos t,3\sin t\rangle

where 0\le t\le\pi. Then the line integral is

\displaystyle\int_Cx^2y\,\mathrm dS=\int_{t=0}^{t=\pi}x(t)^2y(t)\left\|\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\right\|\,\mathrm dt
=\displaystyle\int_{t=0}^{t=\pi}(3\cos t)^2(3\sin t)\sqrt{(-3\sin t)^2+(3\cos t)^2}\,\mathrm dt
=\displaystyle3^4\int_{t=0}^{t=\pi}\cos^2t\sin t\,\mathrm dt

Take u=\cos t, then

=\displaystyle-3^4\int_{u=1}^{u=-1}u^2\,\mathrm du
=\displaystyle3^4\int_{u=-1}^{u=1}u^2\,\mathrm du
=\displaystyle2\times3^4\int_{u=0}^{u=1}u^2\,\mathrm du
=54
6 0
3 years ago
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