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3 years ago

PLS HELP QUICK PLS PLS 13 POINTS

Mathematics

1 answer:

kaheart [24]3 years ago

3 0

C Option 1 is linear because 3% of the initial investment is added to the balance each year.

Option 2 is exponential because the balance is multiplied by 100.2% each month.

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So I have another trigonometry question PLZ HELP ME I'm clueless at this rate I have like a little idea but I'm lost, SO PLZZZ!

Sergio039 [100]

Answer:

1) The windsurfer is approximately 580 ft from each lifeguard stand.

2) The distance between the Earth and Mercury is approximately 61 million miles.

Step-by-step explanation:

The image other two situations described is presented in the attached image to this answer.

1) From the attached image, the windsurfer forms a right angled triangle with each of the lifeguard stand and the midpoint between the two lifeguard stands.

Hence, the angle at the top of the triangle is half of 30°, 15° and the distance from the midpoint of the lifeguard stands to the lifeguard stands is 300/2 = 150 ft.

Let the required distance of the windsurfer from each of the lifeguard stands be x.

Using trigonometric relations,

Sin 15° = (150/x)

x = 150 ÷ sin 15° = 150 ÷ 0.2588

x = 579.6 ft = 580 ft to the nearest whole number

2) From the image, the Sun, Earth and Mercury form a triangle.

Let the possible distance between the Earth and Mercury be y.

Using cosine rule,

y² = 36² + 93² - (2×36×93×cos 22°)

y² = 3,736.5769098208

y = √3,736.5769098208 = 61.13 million miles = 61 million miles to the nearest million miles.

Hope this Helps!!!

5 0

2 years ago

Prove for any positive integer n, n^3 +11n is a multiple of 6

suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0

3 years ago

Read 2 more answers

A person has an unknown element with a volume of 16 cubic inches and a weight of 4.13 pounds.

yawa3891 [41]

Answer is A.Zinc
If it not it still Zinc

7 0

2 years ago

With a headwind a plane traveled 128 km in 2h. The return trip with a tail wind took 24 min less. Find the air speed of the plan

sertanlavr [38]

Based on the information given, it can be noted that the air speed that the plane has will be 71.1 km per hour.

The number of hours used for the return trip will be:

= 2 hours - 24 minutes.

= 1 hour 36 minutes.

= 1.6 hours.

Therefore, the speed of the air plane will be:

= (128 + 128) / (2 + 1.6)

= 256/3.6.

= 71.1 km per hour.

Learn more about speed on:

brainly.com/question/25860159

6 0

2 years ago

On your wedding day you leave for the church 34 minutes before the ceremony is to begin, which should be plenty of time since th

Schach [20]

Answer:

Average speed is 37.35 mi/h

Step-by-step explanation:

given data

leave = 34 minutes before

church distance = 12.0 miles

average speed first 17 minutes = 5.0 mi/h

solution

so we find Total distance travel in first 17 minutes = speed × time

Total distance travel in first 17 minutes = 5 × $\frac{17}{60}$

Total distance travel in first 17 minutes = 1.416 mi

and

Distance Remaining = 12 - 1.416 = 10.584 mi

Time Remaining = 34 - 17 min = 17 min

remaining distance Average speed = $\frac{distance}{time}$

Average speed = $\frac{10.584}{\frac{17}{60}}$

Average speed is 37.35 mi/h

3 0

3 years ago