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adell [148]
3 years ago
13

Helppppp pleaseeeee willlll appreciateddddd

Mathematics
1 answer:
kotegsom [21]3 years ago
3 0
Most likely number 3 
we can guess the average of the two since we don't have the time to actually add all of them and divide and number 3 describes it the most  
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Please help I need it!!!
ziro4ka [17]

Answer:

a

Step-by-step explanation:

it is being reflected of the other end

8 0
2 years ago
Read 2 more answers
In triangle ABC, m∠A=(6x+9)∘, m∠B=(x−8)∘, and the exterior angle at C is 141∘.
Sphinxa [80]

Answer:

12°

Step-by-step explanation:

Let's start with the easy first.

We know that m∠C is 39° because together ∠C and the exterior angle equal 180°. And 180 - 141 = 39.

Now, we can use this to find the remaining two angles.

180° - 39° = 141°

So, this means that we can set up m∠A + m∠B = 141°.

6x + 9 + x - 8 = 141

7x + 1 = 141

7x = 140

x = 20

Finally, we can plug in for our x value and find m∠B.

m∠B = x - 8

m∠B = 20 - 8

m∠B = 12°

3 0
2 years ago
Item 18 You and your friend are selling tickets to a charity event. You sell 7 adult tickets and 16 student tickets for $120. Yo
Jobisdone [24]
$4

You can get this by setting up 2 equations. One for you and one for your friend. Each should have a cost equation. Let x = adult and y = student.

You: 7x + 16y = 120
Friend: 13x + 9 = 140

Solve using any method. 
6 0
3 years ago
Read 2 more answers
Question 15
ycow [4]

The perimeter of the smaller triangle is 8+12+16=36.  The big triangle has perimeter 54.  So our scale factor is 54/36=3/2.  So we scale up the short side, 8, by 3/2, giving our answer:

Answer: 12

8 0
3 years ago
Find the quotient. Write your answer in standard form. : 3-i/3+i
s344n2d4d5 [400]
The answer is <span>C. 4/5 - 3/5 i
</span>
\frac{3-i}{3+i} = \frac{(3-i)(3-i)}{(3+i)(3-i)} \\  \\ &#10;(a+b)(a-b) =  a^{2} -b^{2}  \\  \\ &#10;\frac{(3-i)(3-i)}{(3+i)(3-i)}= \frac{ (3-i)^{2} }{3^{2}-i^{2} } \\  \\ &#10;(a-b)^{2} =a^{2}-2ab+b^{2} \\  \\ &#10; \frac{ (3-i)^{2} }{3^{2}-i^{2} } = \frac{3^{2}-2*3*i+i^{2}  }{3^{2}-i^{2} } = \frac{9-6i+i^{2}}{9-i^{2}}  \\  \\ &#10;i =  \sqrt{-1}  \\ i^{2}=-1 \\  \\ &#10;\frac{9-6i+i^{2}}{9-i^{2}} = \frac{9-6i-1}{9+1} = \frac{8-6i}{10}= \frac{8}{10}- \frac{6i}{10}= \frac{4}{5}- \frac{3}{5}i
8 0
2 years ago
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