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krek1111 [17]
3 years ago
8

Solve each triangle for x.

Mathematics
1 answer:
natulia [17]3 years ago
8 0

Answer:

56 deg

Step-by-step explanation:

All internal angles of a triangle add up to 180 degrees

hence,

∠O + ∠R + ∠F = 180

x + (x+43) + (x-31) = 180

x + x+43 + x -31 = 180

3x+ 12 = 180

3x = 180 - 12 = 168

x = 168/3 = 56 deg

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Can I please get some help with both problems
dimulka [17.4K]
So you subtract 180 from 90 you get you answer then you will divide it by 55 and your remaining answer which you will get will be x 
4 0
3 years ago
Note that the right triangles with sides equal to 5, 12, 13 and 9, 12, 13 both have a side equal to 12. using this fact find the
kondor19780726 [428]
There is a mistake 9,12,13 is not a right triangle. but 9,12,15 is. IF you use this triangle instead, then the area of the 13,14,15 triangle is the sum of the area of the other two triangles.
6 0
4 years ago
Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.
dusya [7]

The given equation has no solution when K is any real number  and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

<h3>What is the condition for a solution?</h3>

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

To learn more about the solution visit:

brainly.com/question/1397278

5 0
2 years ago
Read 2 more answers
On Friday, 80 children and 48 adults went to a movie. What was the ratio of children to adults? A 5 to 3 B 10 to 4 C 6 to 3 D 2
Setler79 [48]
A .  Because one you make the ratio 80:48 you simplify (divide until you can divide anymore ).
7 0
3 years ago
Read 2 more answers
A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.
Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean(\mu) = 500 and a standard deviation(\sigma) =  100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = \dfrac{Y-\mu}{\sigma}

Z-score = \dfrac{625-500}{100}

⇒ Z-score = \dfrac{125}{100}

⇒Z-score =1.25

Therefore , the required z-score = 1.25

6 0
3 years ago
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