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Ilia_Sergeevich [38]
3 years ago
11

QS¯¯¯¯¯¯¯ bisects ∠PQR . If m∠PQS = 5x and m∠RQS = 2x + 6, then what is m∠PQR ? m∠PQR is

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

PQR = 20

Step-by-step explanation:

Given

Bisector: QS

PQS = 5x

RQS = 2x + 6

Required

Determine PQR

Since PQR is bisected to PQS and RQS, we have that

PQS = RQS

Substitute expressions for PQS and RQS

5x = 2x + 6

Collect Like Terms

5x - 2x = 6

3x = 6

Divide both sides by 3

x = 2

Solving for PQR;

PQR = PQS + RQS

PQR = 5x + 2x + 6

PQR = 7x + 6

Substitute 2 for x

PQR = 7 * 2 + 6

PQR = 14 + 6

PQR = 20

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Step-by-step explanation:

6 0
3 years ago
Using principle of mathematical induction prove that 6^-1 divisble by 5 .​
salantis [7]

I suppose the claim is 5 \mid 6^n - 1 for n\in\Bbb N.

When n=1, we have 6^1 - 1 = 6 - 1 = 5, and of course 5 divides 5.

Assume the claim holds for n=k, that 5 \mid 6^k - 1. We want to use this to show it holds for n=k+1, that 5 \mid 6^{k+1} - 1.

We have

6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5

Since 5 \mid 6^k - 1, we can write 6^k - 1 = 5\ell for some integer \ell. Then

6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)

which is clearly divisible by 5. QED

5 0
2 years ago
If the probability of an event happening is p, what is an algebraic expression that the event does not happen?
inna [77]
H=nh is a possible algebraic expression that the event doesn't happen. Hope this helps!
8 0
3 years ago
I REALLY NEED HELP ON THIS PLS HELP
Eva8 [605]

Part (a)

Answer: See the attached image below to see the filled out chart.

Note how rational and irrational numbers have nothing in common. This means there is no overlap. So they go in the rectangles. The two sets of numbers join up to form the entire set of real numbers.

Integers are in the set of rational numbers. This is because something like 7 is also 7/1; however 1/7 is not an integer. So not all rational numbers are integers. The larger purple circle is the set of integers.

The smaller blue circle is the set of whole numbers. The set of whole numbers is a subset of integers. Recall the set of whole numbers is {0,1,2,3,...} so we ignore the negative values only focusing on 0 and positive numbers that don't have any fractional values. In contrast, the set of integers is {..., -3, -2, -1, 0, 1, 2, 3, ...} here we do include the negatives.

============================================================

Part (b)

1) Some irrational numbers are integers.

This is false. An irrational number is not rational. The set of integers is contained entirely in the set of rational numbers.

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

2) Some whole numbers are not irrational numbers.

This is false. The statement implies that some whole numbers are irrational, but the set of whole numbers is inside the set of rational numbers, which has no overlap with the irrationals. The statement should be "All whole numbers are not irrational numbers".

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

3) All rational numbers are whole numbers.

This is false. A rational number like 1/3 is not a whole number.

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

4) All integers are whole numbers

This is false. An integer like -44 is not a whole number because the set of whole numbers is {0,1,2,3,...} and we're not including negative values.

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

In summary, all four statements for part (b) are false.

8 0
3 years ago
Read 2 more answers
A rowing team rowed 90 miles while going with the current in the same amount of time as it took to row 10 miles going against th
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Answer:   
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 .. 
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 80x=400 
 x=5  rate of the rowing team in still water=5 mph

4 0
3 years ago
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