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kiruha [24]
3 years ago
5

Which fraction is equivalent to 4/7?

Mathematics
1 answer:
forsale [732]3 years ago
6 0

Answer: Depends....could be anything...one example is 8/14

Step-by-step explanation:

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How many different ways can a coach select the captain and co-captain of a team from a group of 20 people?
sergij07 [2.7K]
The coach can select the captain in 20 ways (i.e. from all the available 20 players), and he can select the co-captain in 19 ways (i.e. after selecting the captain, there will be 19 people left).
Therefore, the coach can select the captain and the co-captain in 20 x 19 = 380 ways.
5 0
3 years ago
Read 2 more answers
PLEASE ANSWER Chelsea is budgeting for her trip to the mall. She does not want to spend any more than $140. If she wants to buy
Tcecarenko [31]
Let s equal the number of shirts she can buy

140 > $28.50 + $20.75s
111.50 > 20.75s
5.37 > s

so we round down.  she can buy 5 shirts, plus the one dress and she would spend less than $140

4 0
3 years ago
Save and
Darya [45]

Answer: 6/5

Step-by-step explanation:

7 0
2 years ago
Given: △ACM, m∠C=90°, CP ⊥ AM
larisa86 [58]

Answer:  The answer is 3\dfrac{4}{7}.

Step-by-step explanation:  Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.

Now,

AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)

Comparing equations (A) and (B), we have

5x-1=7x^2-1\\\\\Rightarrow 5x=7x^2\\\\\Rightarrow x=\dfrac{5}{7},~\textup{since }x\neq 0.

Thus,

AM=5\times\dfrac{5}{7}=\dfrac{25}{7}=3\dfrac{4}{7}.

8 0
3 years ago
Pls help only b,d, and f
Leya [2.2K]
Solving for y right?

2y = 3x + 4 - x
2y = 2x + 4
y = 2x + 4 over 2
y = 2 (x + 2) over 2
y = x + 2


y - 3 = 2x - 6 over 2
y - 3 = 2(x - 3) over 2
y - 3 = x - 3
y = x


x - y - 2 = 2(2x + 1)
-y - 2 = 2(2x + 1) - x
-y = 2(2x + 1) - x + 2
y = -2(2x + 1) + x - 2
8 0
3 years ago
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