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dedylja [7]
3 years ago
15

What is the length of leg s of the triangle below?

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

the answer is F.8

Step-by-step explanation:

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Find the surface area of a cone with slant height of 9 inches and a radius of 3 inches. ANSWERS: 117 in2 81 in2 109 in2 113 in2
oee [108]

Answer:117in2

Step-by-step explanation:

7 0
3 years ago
The coordinates of the vertices of a polygon are shown below.
MatroZZZ [7]

Answer: Hexagon

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Step-by-step explanation: A six-sided shape is a hexagon, a seven-sided shape a heptagon, while an octagon has eight sides… There are names for many different types of polygons, and usually the number of sides is more important than the name of the shape. There are two main types of polygon - regular and irregular.

6 0
2 years ago
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100 POINTS 10 QUESTIONS PLEASE HELP PICTURES BELOW PLEASE SPECIFY WHICH QUESTION YOU ARE ANSWERING
Semenov [28]

Answer:

The answers are \frac{2}{5}, \frac{1}{4}, \frac{1}{10}, \frac{5}{2}, \frac{5}{8}, \frac{4}{1}, \frac{8}{5}, and \frac{10}{1}.

Step-by-step explanation:

Proportions are fractions that can be made by using the given numbers, which in this case are 2, 5, 8, and 20. Let's pair each one with the other three and then simplify if possible.

First, let's begin with 2:

\frac{2}{5} = \frac{2}{5}

\frac{2}{8} = \frac{1}{4}

\frac{2}{20} = \frac{1}{10}

Then, let's do 5:

\frac{5}{2} = \frac{5}{2}

\frac{5}{8} = \frac{5}{8}

\frac{5}{20} = \frac{1}{4}

Note that we already have \frac{1}{4}, so we do not need to include an additional one.

Now, let us do 8:

\frac{8}{2} = \frac{4}{1}

\frac{8}{5} = \frac{8}{5}

\frac{8}{20} = \frac{2}{5}

See how we already have \frac{2}{5}, so we won't have to include that as well.

Finally, let's do 20:

\frac{20}{2} = \frac{10}{1}

\frac{20}{5} = \frac{4}{1}

\frac{20}{8} = \frac{5}{2}

Now see that we already have both \frac{4}{1} and \frac{5}{2}, so we won't have to include both of them, as they are both extras.

Hence, the answers are \frac{2}{5}, \frac{1}{4}, \frac{1}{10}, \frac{5}{2}, \frac{5}{8}, \frac{4}{1}, \frac{8}{5}, and \frac{10}{1}.

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
8 0
3 years ago
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Yesterday, between noon and midnight, the temperature dropped by 25.8°F. If the
SVEN [57.7K]

15.1 f

Step-by-step explanation:

-13.4+25.8= 15.1 F

8 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
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