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Ann [662]
3 years ago
11

How many triangles can be constructed with Three angles each measuring 60 degrees

Mathematics
1 answer:
nalin [4]3 years ago
5 0
If each angle is 60 you can only make one triangle.
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Plz help before 11:59 pm I will mark brainliest
olganol [36]

Answer:

1.- 140 in²

2.- 2.25 in²

3.- 43.25 in²

4.- 7.5 in²

5. 349.75 in²

Step-by-step explanation:

4 congruent triangles mean 4 triangles same measurement

area 1 triangle=(B*H)/2=(10*7)/2=35in²

for area of the roof are 4 by area 1 triangle=4*35=140 in²

2.- area hole. Hole is a square so area Is side by side

area=1.5*1.5=2.25

3.- area front except the hole. As the side of front is isosceles trapezoide which it’s area is (B1+B2)H/2 where B1=8 in and B2=5 in and H=7 in

so arera with hole=(8+5)*7/2=45.5 in²

7 0
3 years ago
Read 2 more answers
Jim earns $1,600 per month after taxes. He is working on his budget and has the first three categories finished.
ivann1987 [24]
<span>He is budgeting too much for transportation. He has used the highest recommended percentages to calculate the amounts for the three categories. He is budgeting too much for housing. He has used the lowest recommended percentages to calculate the amounts for the three categories. hope this helps :)</span>
6 0
3 years ago
Read 2 more answers
Write each improper fraction as a mixed number: 93/12
yuradex [85]

Answer:

The answer would be 7 3/4

Step-by-step explanation:

4 0
3 years ago
Does anybody know anything about simplifying rational expressions? 4x/11x
babunello [35]
All you have to do is cancel the x 4/11
7 0
3 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

5 0
2 years ago
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