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elena55 [62]
3 years ago
9

Which is equivalent to the series shown?

Mathematics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

please I don't know

Step-by-step explanation:

please I don't know

please give me another question

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Gavyn and Alex win some money and share it in the ratio 4:1. Gavyn gets £18 more than Alex. How much did they get altogether?
densk [106]

Answer:

£30

Step-by-step explanation:

Gavyn has £18 more than Alex

—› G - A = £18

—› 4u - 1u = 3u = £18

1u = £18 ÷ 3

= £6

total units = 4 + 1 = 5

5u = 5 × £6 = £30

3 0
3 years ago
Twenty less than thrice a number is 223.what is the number​
Tomtit [17]

So I just did 3x-20=223

3x=243

x=243/3

x=81

so the answer would be 81

8 0
3 years ago
Read 2 more answers
Find the surface area of the part of the circular paraboloid z=x^2+y^2 that lies inside the cylinder X^2+y^2=1
hichkok12 [17]

Answer:

\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}

Step-by-step explanation:

Given that:

The surface area (S.A) z = x^2 +y^2

Hence the S.A is of form z = f(x,y)

Then the S.A can be represented with the equation

A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA

where :

D = cylinder x^2 +y^2 =1

In polar co-ordinates:

D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)

Similarly, \dfrac{\partial z}{\partial x} = 2x and \dfrac{\partial z}{\partial y} = 2y

Therefore;

S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA

= \iint_D \sqrt{1+4(x^2+y^2)} \ dA

= \int^{2 \pi}_{0} \int^{1}_{0}  \sqrt{1+4r^2} \ r \ dr \d \theta

= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0

= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]

\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}

6 0
3 years ago
Place the following rates of change in order<br> from least to greatest:)
Assoli18 [71]

Answer:

-4 , 0 , 4/5 , 90% , 2

Step-by-step explanation:

-4 , 0 , 0.8 , 0.9 , 2

5 0
3 years ago
A football player kicks a football and records the height of the ball at different times.
Ronch [10]

show to your tcher.....

4 0
3 years ago
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