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Mariulka [41]
3 years ago
7

Write a variable expression to describe the rule for the sequence. Then find the 100th term in the sequence.

Mathematics
1 answer:
vladimir1956 [14]3 years ago
7 0

f(x)=x+5 so the 100th would be 105

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Ion 1(Multiple Choice Worth 2 points)
Alexxandr [17]

Answer:

105.19.

Step-by-step explanation:

Area = 1/2 ac sin B

= 1/2 * 10 * 22 * sin 107

= 105.19..

4 0
3 years ago
First person to answer gets Brainliest and 100 points even though you get it for answering. No nonsense answers please.
Artemon [7]

Answer:

Hi, I am little confused for what im supposed to do can you please help me understand a little more so i i can answer, thanks!

Step-by-step explanation:

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3 years ago
Write the coordinates of point a. please help?
Over [174]
The answer for point A is (-4,1)
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Find three solutions for this linear function and choose the correct graph by clicking on it. F -1(x) if F(x) = x + 3
algol13

Answer:

[1,2], [2,1], [3,0]

Step-by-step explanation:

To find F-1(x) if F(x)=x+3 we just need to put -x instead of x.

From our graph(attached) we multiply any three solutions by a -1:

F-1(x)\\\\#For \ x=1\\F-1(1)=-1+3=2\\\\#For \ x=2\\F-1(2)=-2+3=1\\\\#For \ x=3\\F-1(3)=-3+3=0

Hence, the 3 solutions are [1,2],[2,1],[3,0]

3 0
3 years ago
A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
qaws [65]

The projectile's horizontal and vertical positions at time t are given by

x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t

y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2

where g=9.8\dfrac{\rm m}{\mathrm s^2}. Solve y=0 for the time t it takes for the projectile to reach the ground:

30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s

In this time, the projectile will have traveled horizontally a distance of

x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m

The projectile's horizontal and vertical velocities are given by

v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ

v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about \sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}.

7 0
3 years ago
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