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JulijaS [17]
3 years ago
6

The basketball team has 50 members, and 18 of them are 6th graders. What percentage of the team is in the 6th grade?​

Mathematics
2 answers:
motikmotik3 years ago
8 0

Answer:

36%

Step-by-step explanation:

\frac{18}{50} = \frac{36}{100}

So 36% of the team is in 6th grade.

Ostrovityanka [42]3 years ago
3 0

<em>Answer: </em><u><em>36%</em></u><em>. one hundred divided by two equals fifty, eighteen multiplied by two equals </em><u><em>36%</em></u><em>, therefore, its </em><u><em>36%</em></u>

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In the right triangle ?ABC, leg AC=6 cm and leg BC=8 cm. Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5. Find area of ?MN
Licemer1 [7]

Given : In Right triangle ABC, AC=6 cm, BC=8 cm.Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5.

To find : Area (ΔMNC)

Solution: In Δ ABC, right angled at C,

AC= 6 cm, BC= 8 cm

Using pythagoras theorem

AB² =AC²+ BC²

      =6²+8²

     = 36 + 64

→AB²  =100

→AB²  =10²

 →AB  =10

Also, AM:MN:NB=1:2.5:1.5

Then AM, MN, NB are k, 2.5 k, 1.5 k.

→2.5 k + k+1.5 k= 10

→ 5 k =10

Dividing both sides by 2, we get

→ k =2

MN=2.5×2=5 cm, NB=1.5×2=3 cm, AM=2 cm

As Δ ACB and ΔMNC are similar by SAS.

So when triangles are similar , their sides are proportional and ratio of their areas is equal to square of their corresponding sides.

\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}

\frac{Ar(ACB)}{Ar(MNC)}=4

But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]

\frac{24}{Ar(MNC)}=4

→ Area(ΔMNC)=24÷4

→Area(ΔMNC)=6 cm²

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3 years ago
Sally's job pays her $3,500 per month. What is her gross average weekly wage?
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6 0
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Aleks04 [339]

r and s are parallel

given ∠4 = ∠5 ( these are congruent interior alternate angles )

between the lines r and s, hence

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4 0
3 years ago
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