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Aleksandr-060686 [28]
3 years ago
9

Find the measures of the missing angles

Mathematics
1 answer:
rodikova [14]3 years ago
7 0
2,7, and 6 are all 70 degrees. 1,4,5, and 8 are all 110 degress
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If x−2/4=2, then x=10
lara31 [8.8K]

Answer:

True.

Step-by-step explanation:

\frac{(10)-2}{4}=2\\\\\frac{8}{4}=2\\\\2=2

8 0
3 years ago
Read 2 more answers
What is the perimeter of the entire football field including the end zone
Dmitry [639]

Answer:

346 2/3 yards

Step-by-step explanation:

A football field is 100 yards long with each end zone being 10 years long, meaning the total length is 120 yards.

The width is 53 1/3 yards.

The perimeter is all the sides added, and a rectangle's, the shape of the field's, formula is p = 2l + 2w.

So if we plug it in and add it all up:

120 + 120 + 53 1/3 + 53 1/3

Perimeter = 346 2/3

4 0
3 years ago
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Haddie flipped a fair coin 12 times. The coin landed on heads 11 times and tails 1 time. She flips the coin a 13th time. How lik
Firdavs [7]

Answer:

1 out of 2

Step-by-step explanation:

even though the coin was landing on heads more times than tails it still remains a fifty-fifty chance.

7 0
2 years ago
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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
What value of x is in the solution set of –5x – 15 > 10 + 20x?
Scilla [17]
Hello,

-5x-15>10+20x
==>-15-10>20x+5x
==>-25>25x
==>-1>x
==>x<-1

6 0
3 years ago
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