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antiseptic1488 [7]
3 years ago
8

What is 7/8 divided by 11/16

Mathematics
2 answers:
Igoryamba3 years ago
8 0
1.27 repeated is the answer
insens350 [35]3 years ago
3 0
7/8 = 0.875 11/16 = 0.6875 0.875 ÷ 0.6875 = 1.27 or 1.3
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What is an equation of the line that passes through the point (2,−6) and is parallel to the line x−2y=8?
lorasvet [3.4K]

Answer:

y = x/2 - 7

Step-by-step explanation:

First, we need to find the slope of the given equation: x - 2y = 8

Subtract x from both sides

x - 2y = 8

- x        - x

-2y = 8 - x

Divide both sides by -2

-2y/-2 = (8 - x)/-2

y = -4 + x/2

The slope of this equation is 1/2

So the equation of our parallel equation is y = x/2 + b

We have to find b, so plug in the given coordinates

-6 = 2/2 + b

-6 = 1 + b

Subtract 1 from both sides

-6 = 1 + b

- 1    - 1

b = -7

Plug it back into the original equation

y = x/2 - 7

6 0
2 years ago
Find the midpoint of a and b when a has coordinates (-5,3) and b has coordinates (3,-1)
BlackZzzverrR [31]

Answer:

The midpoint is (2 , 1)

Step-by-step explanation:

To find the midpoint, we have to add the corresponding coordinates

[(-5 , 3) + (3 , -1)] / 2

we separate into the corresponding

(-5 + 3) / 2 =

-2 / 2 = -1

(3 - 1) / 2 =

2 / 2 = 1

The midpoint is (2 , 1)

3 0
3 years ago
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The perimeter of the a and BC is 14 unitsBecause if you add all of the sides together you get 14 and I need points
6 0
3 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
Joshua went kayaking. It cost him $ 15 to rent the kayak. It also cost $ 6 for each hour spent on the lake kayaking. Joshua spen
pishuonlain [190]

Step-by-step explanation:

The total bill will be the sum of the rent and the cost for the total time spent.

$15 + $6 * 2.5

$15 + $15

$30

8 0
2 years ago
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