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STALIN [3.7K]
3 years ago
9

Find the length of y. assume the triangles are similar

Mathematics
2 answers:
Talja [164]3 years ago
3 0

Hey there! I'm happy to help!

If the triangles are similar, the ratio between all of the sides is the same.

We see that to get the 4.2 in the big triangle to the 2.4 in the smaller triangle, we must divide by 1.75 (we get this from 4.2/2.4=1.75). So, if the ratio between all the sides is the same in both triangles, we can divide 6.3 by 1.75 to get y.

6.3/1.75=3.6

Therefore, the length of y is B. 3.6.

Have a wonderful day! :D

Natasha_Volkova [10]3 years ago
3 0

Answer:

3.6 = y

Step-by-step explanation:

We can use ratios to solve

2.4             y

----------- = ------------

4.2              6.3

Using cross products

2.4 * 6.3 = 4.2 y

15.12=4.2y

Divide by 4.2

15.12/4.2 = 4.2y/4.2

3.6 = y

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Read 2 more answers
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
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Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
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