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lana [24]
2 years ago
5

If you launch the cannonball from the ground instead (hi=0), what does that change in the original equation?

Mathematics
1 answer:
ikadub [295]2 years ago
7 0

Answer:

  the value of hi changes from 58 to 0:

  h(t) = -4.9t^2 +19.8t

Step-by-step explanation:

If the initial height changes, then the term in the equation that represents that height will change. The initial height term is hi, so it changes from 58 to 0.

  h(t)=-4.9t^2+19.8t+0\\\\\boxed{h(t)=-4.9t^2+19.8t}

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A cube has a volume of 2197 m^3. Its surface is to be painted. Each can of paint covers about 40 m^2. How many cans of paint are
Gelneren [198K]
<em>Volume= a³</em>
<em>a=edge</em>
<em>2197 = a³</em>
<em>=> a=13 m</em>
<em>Surface area=6a²=6*13²=6*169=1014 m²</em>
<em>\frac{1014}{40} = 25,35 ≈ 26</em>
<em>=> <u>26 cans needed</u></em>
8 0
3 years ago
Will purchased a $175,000 home with a 7/23 balloon mortgage. His initial rate was 3.5%. At the end of the initial rate, he decid
Ganezh [65]

Answer: $798.94

Step-by-step explanation:

8 0
3 years ago
Find the sum or difference. a. -121 2 + 41 2 b. -0.35 - (-0.25)
s344n2d4d5 [400]

Answer:

2

Step-by-step explanation:

The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at the sequence of numbers

1

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

etc.,

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + · · ·

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + · · ·

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + · · ·

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + · · ·

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + · · · + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + · · · + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, . . . , a r^n all cancel and you are left with S - r S = a - a r^(n+1), so

(IMAGE)

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

8 0
3 years ago
Jose just submitted his resume to a job posting. The posting started thats the job pays $47.588 annually. What would Jose’s mont
Marrrta [24]

Answer:

3965.67 a month

Step-by-step explanation:

presuming you mean $47,588 annually.

47588/12

8 0
3 years ago
I need some help with these two questions please help!
antoniya [11.8K]

Answer:

Step-by-step explanation:

-10.6*0.5=m+11.7

-5.3=m+11.7

-5.3-11.7=m

-17=m

14.2=2(-5.8+t)

14.2=-11.6+2t

14.2+11.6=2t

25.8=2t

25.8/2=t

12.9=t

6 0
3 years ago
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