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Mariana [72]
3 years ago
7

Points J and K are midpoints of the sides of triangle FGH. What is the value of y?

Mathematics
2 answers:
sashaice [31]3 years ago
7 0
The answer 

the complement of the question is 

What is the value of y?

2
5
7
<span>8
</span>
According to the image, HKJ and FGH are similar, we can apply the theorem of thales
since GK and GF are parallels, so 

HK / HF = HJ / HG = KJ / GF and as we see on the figure, HK = HF /2, this implies  2HK =HF

HK / 2HK =  KJ / GF  it is equivalent to   1/2 = 2y+5 / 5y +3 

likewise, equivalent to  5y +3  = 4y +10, 5y-4y = 10-3, and y = 7


so the answer is 7




Oksanka [162]3 years ago
7 0

Answer:7

Step-by-step explanation:

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Bryan has a square parcel of land that is 120 yards long and 120 yards wide. The land is planted with Bermuda grass and surround
Mekhanik [1.2K]
The diagonal fencing will be the hypotenuse of a right triangle with sides 120yds long so:

h=√2(120^2))

h=√28800 yd

And the surrounding fencing will just be four times the side length of 120yds so:

s=480yd

So the total fencing needed is:

f=480+√28800 yards

f≈649.71 yd  (to nearest hundredth of a yard)  which is

f≈650 yds  (to nearest whole yard)

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f=√28800

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6 0
3 years ago
Read 2 more answers
Which of the following is a prime number 67,65,63 or 60
sesenic [268]
Use the rule of divisibility :
60 is composite because it is even so it can be divided by 2
67 is prime because none if the divisiblility rules work on it
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7 0
4 years ago
I need answer Immediately pls!!!!!!
Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

n(U)=27

n(A)=7

n(B)=18

n(A'\cap B')=7

We know that,

n(A\cup B)=n(U)-n(A'\cap B')

n(A\cup B)=27-7

n(A\cup B)=20

Now,

n(A\cup B)=n(A)+n(B)-n(A\cap B)

20=7+18-n(A\cap B)

n(A\cap B)=7+18-20

n(A\cap B)=25-20

n(A\cap B)=5

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}

\text{Probability}=\dfrac{5}{27}

Therefore, the required probability is \dfrac{5}{27}.

3 0
3 years ago
During basketball practice, Grant made 8 out of 10 free throws. If he attempts 50 free throws, how many times is he likely to mi
Murrr4er [49]
Hi 

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8 0
3 years ago
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insens350 [35]

Answer:

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