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kipiarov [429]
3 years ago
12

HELLLLLLP!!!!

Mathematics
1 answer:
mixer [17]3 years ago
7 0
Do $2.689 - $2.569 which would yield an increase of 12 cents
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Use the diagram. AB is a diameter, and AB is perpendicular to CD. The figure is not drawn to scale.
Westkost [7]

Answer:

The measure of arc BD is 147°

Step-by-step explanation:

we know that

If segment AB is perpendicular to segment CD

then

The measure of the inner angle CPA is a right angle (90 degrees)

Remember that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

so

m∠CPA=(1/2)[arc AC+arc BD]

substitute the given values

90°=(1/2)[33°+arc BD]

180°=33°+arc BD

arc BD=180°-33°=147°

7 0
2 years ago
A bag contains 50 lottery balls, numbered 1-50. A ball is chosen at random, then a month of the year is selected. What is the pr
Morgarella [4.7K]

Answer:

4/75

Step-by-step explanation:

We have two sample spaces

1.Lottery balls

2. Month of the year

For lottery balls the sample space is S= 50

Multiples of 3 between 1 and 50 are

(3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48)= 16 in numbers

Hence the probability of choosing a multiple of 3

Pr(multiple of 3) = 16/50= 8/25

Also the sample space for months of the year is S= 12

Two months starts with letter m, March and may

Pr(of months with m) = 2/12= 1/6

the probability of choosing a multiple of 3 and a month starting with the letter M= 8/25*1/6= 8/150= 4/75

6 0
3 years ago
PLEASE HELP WILL MARK BRAINLIEST!!
Gre4nikov [31]

Answer:

Explanation below.

Step-by-step explanation:

1. a. \sqrt{x} =12

  b. \sqrt{x} ^{2}=12^{2}

  c. x=144

2. a. \sqrt{x^{2} } =\sqrt{196}

   b. x= 14

3.  

\sqrt{x} -4 +4= 8+4\\\sqrt{x} =12\\x= 3.46410161514  \\or  \\2\sqrt{3}

4.

2(x^{2} +3)=398\\x^{2} +3= 199\\x^{2} =196\\x=14

7 0
3 years ago
Read 2 more answers
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
22/58 simplified pleas herry its timed
Alex

Answer:

11/29

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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