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qwelly [4]
3 years ago
9

Solve the word problem. The formula s = gives the side length s of a regular pentagon with perimeter P. What is the side length

of a regular pentagon with perimeter 7.5 feet? A. 1.25 ft B. 1.5 ft C. 2.5 ft D. 12.5 ft
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0
He side length would be 1.5ft.
Hope This Helps and God Bless!
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HELP PLEASE !!29. How do the graphs of the functions y = f(x) and y = f(–x) relate when
Nezavi [6.7K]
The answer for 29 is A
5 0
3 years ago
X^4 a^5 / x^3 a^3<br><br> answers<br><br> x^2 a^2<br> xa^2<br> x^-1 a^-2<br> a^2/x
bagirrra123 [75]
Well the answer i got was a^{8}x when i multiplied
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3 0
3 years ago
Is this correct now?
Makovka662 [10]
I would say no it’s not correct, the point is not in the correct place
8 0
3 years ago
Find the sum of 1 + 3/2 + 9/4 + …, if it exists.
zmey [24]

Answer:

Option (4)

Step-by-step explanation:

Given sequence is,

1+\frac{3}{2}+\frac{9}{4}..........

We can rewrite this sequence as,

1+\frac{3}{2}+(\frac{3}{2})^2.............

There is a common ratio between the successive term and the previous term,

r = \frac{\frac{3}{2}}{1}

r = \frac{3}{2}

Therefore, it's a geometric sequence with infinite terms. In other words it's a geometric series.

Since sum of infinite geometric sequence is represented by the formula,

S_{n}=\frac{a}{1-r}  , when r < 1

where 'a' = first term of the sequence

r = common ratio

Since common ratio of the given infinite series is greater than 1 which makes the series divergent.

Therefore, sum of infinite terms of a series will be infinite Or the sum is not possible.

Option (4) will be the answer.

3 0
4 years ago
Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.
andrey2020 [161]
Check the picture below.

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keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

\bf \textit{parabola vertex form with focus point distance}\\\\&#10;\begin{array}{llll}&#10;(y-{{ k}})^2=4{{ p}}(x-{{ h}})&#10;\\\\&#10;\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ({{ h}},{{ k}})\\\\&#10;{{ p}}=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;h=4\\&#10;k=1\\&#10;p=4&#10;\end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1)&#10;\\\\\\&#10;\cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y

8 0
4 years ago
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