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Oksanka [162]
3 years ago
8

Show problem set up for full credit.

Mathematics
1 answer:
ikadub [295]3 years ago
6 0

Answer:

If the arcs are the same

4x+25=5x

x=25

So, QPR is 4(25)+25

QPR is 125

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Which set of integers is NOT a Pythagorean triple and are NOT side lengths of a right triangle?
Arlecino [84]

Answer:

Answer is 27, 32, 45

Step-by-step explanation:

In the right angle triangle, hypontenous is the longest side.

∴ In each option square of longest side has to be equal to the sum of square of other two side.

We know, h^{2} = a^{2} + b^{2}  ( as per pythogorean theorem)

If we check the last option; 27, 32, 45

In the given set of integer, we have longest side as 45

∴ conisdering 45 as hyptenous

Subtituting the value in the formula

⇒45^{2} = 27^{2} +32^{2}

⇒2025= 729+1024

⇒2025\neq 1753

∴ LHS\neq RHS

Hence, the set of given integer is not a pythagorean triple and are not side length of right angle.

5 0
3 years ago
Read 2 more answers
Solve each system by elimination:<br> 10x+8y= -18 <br> 8x+8y= -24
diamong [38]

Answer:

Step-by-step explanation:

10x + 8y = -18

-8x - 8y = 24

2x = 6

x = 3

8(3) + 8y = -24

24 + 8y = -24

8y = -48

y = -6

(3, -6)

7 0
3 years ago
Read 2 more answers
1. Parker paid $4.50 for three pounds of gummy candy. Assuming each pound of gummy candy costs the same amount, complete the tab
Lunna [17]

Answer:

1: $1.50

2: $3.00

4: $6.00

5: $7.50

6: $9.00

7: $10.50

8: $12.00

9: $13.50

Step-by-step explanation: To answer this problem you need to divide $4.50 by 3. When you do that you get $1.50 per pound of gummy bears. Then you can either add $1.50 to each amount or multiple them by the x value. Then graph the numbers you get.

4 0
3 years ago
Given f(x)=x-10 . what is f(-7)?
Tomtit [17]
F(-7) = -7 - 10

f(-7) = -17


Hope I helped!

Let me know if you need anything else!

<span>~ Zoe (Rank:'Genius')</span>
8 0
3 years ago
Let A, B and C be sets. Prove that if A ⊆ B ∪ C and A ∩ B = ∅ then A ⊆ C
Studentka2010 [4]

Answer and Solution:

As per the question:

Given:

If A\subseteq B\cup C

A\cap B = \phi

To prove:

A\subseteq \C

Proof:

Suppose t\in A

As we know that:

A\subseteq B\cup C

Therefore,

t\in B or t\in C

Now, if we assume that t\in B

Then

t\in A\cap B

Since,

t\in A and t\in B

But

A and B are disjoint set and A\cap B = \phi

Therefore, this is contradictory.

Thus

t\notin B

So,

t\in C

Every element in the set A is also present in the set C

Therefore, A\subseteq \C

Hence, proved.

6 0
4 years ago
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