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bija089 [108]
3 years ago
7

An astronomy club wants to buy six new eyepieces for their telescope. Four of the lenses cost $53.98 each. Two lenses cost $21 e

ach. What is the total cost?
Also, please show your work and tell me how you got that answer.
Mathematics
2 answers:
snow_lady [41]3 years ago
7 0

Answer:

239.92

Step-by-step explanation:

53.94×4=215.92

Then you do 21×2=24 then you take the 215.92 and the 24 and add them together and it gives you 239.92

dexar [7]3 years ago
3 0

The magnification of a telescope is the ratio of its focal length to the focal length of the eyepiece in use. Thus a telescope with a 1000 mm focal length, used with an eyepiece of 25 mm focal length, has a magnification of 1000 / 25, or 40. It makes things look 40 times wider, or if you prefer, 40 times closer.

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2 years ago
100 POINTS HELP ASAP NOW!!!!!!<br> Find the value of x in each case.
Zepler [3.9K]

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Answer:

  x = 22.5°

Step-by-step explanation:

The interior angle at E is (180 -4x), the supplement of the exterior angle there, 4x. The sum of angles 2x and (180-4x) will be equal to 6x, because alternate interior angles at transversal BE are congruent:

  2x +(180 -4x) = 6x

  180 = 8x . . . . . . . add 2x and simplify

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5 0
3 years ago
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Marina86 [1]
Hey man I got u. The answer for this question is B. Peace
3 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%287%20-%208%29%20%5Ctimes%202%20%7B%7D%5E%7B2%7D%20" id="TexFormula1" title="(7 - 8) \times 2
Diano4ka-milaya [45]
Answer: -4
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7 0
3 years ago
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1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle
zimovet [89]

Answer: \left |  2t-5\right |,\ 2,\ 2t-5

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

s(t)=t^2-5t+1

Speed of the particle is given by

\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left |  2t-5\right |

Acceleration of the particle is

\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2

velocity can be negative, but speed cannot

\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5

3 0
3 years ago
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