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3 years ago

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Question 7

1 answer:

Natali5045456 [20]3 years ago

4 0

To solve take circumference of the lamp shade( 2pi(r) or pi(d) ) so circumference is 12pi.
Then multiple by hight or altitude to get surface area ( circumference x height)
SA =72pi or ~226.19 in^2

If you are putting material on the top of lamp to, add pi(r)^2 or 36pi= ~113.10 in^2

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15r-15s=<br> can you help

Alla [95]

Answer:

i dlnt know 15 I think not sure

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3 years ago

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Which point on the x-axis lies on the line that passes through point P and is perpendicular to line MN? (0, 1) (0, 4) (1, 0) (4,

LenaWriter [7]

I need to see line MN and point P first on a graph.

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3 years ago

Patrick is driving form Sydney to Canberra. He travels for 288km trip in 4 hours. What's his average speed?

nika2105 [10]

Answer:

v = 72 km/h

Step-by-step explanation:

Given that,

The distance covered by the Patrick, d = 288 km

Time taken, t = 4 hours

We need to find the average speed of Patrick. We know that the average speed of an object is equal to the total distance covered divided by total time taken. Let it is v. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{288\ km}{4\ h}\\\\v=72\ km/h$

So, his average speed is equal to 72 km/h.

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3 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

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3 years ago

Evaluate x-(-y) when x= -2.31 and y = 5.9

kondor19780726 [428]

-2.31-(-5.9)= -2.31+5.9= 3.59

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3 years ago

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