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telo118 [61]
3 years ago
10

A group of 160 people is going on a boat tour. If each boat holds 8 people. How many boats will they need?

Mathematics
1 answer:
Llana [10]3 years ago
5 0
20 people would go on each boat because 160 divided by 8 is 20

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Frank is making a pennant in the shape of a triangle for his senior class photo. He wants the base length of this triangle to be
Sav [38]

Answer:

Height should be ≤ 9 inches.

Step-by-step explanation:

Given:

Frank is making a pennant in the shape of a triangle for his senior class photo.

Base of triangle = 6 in

Area of triangle ≤ 27 in^2

Let height of the triangle be h

Now we now that,

Area of triangle = \frac{1}{2}\times base \times height

\frac{1}{2}\times 6 \times h \leq 27\\\\3h\leq 27\\\\h\leq \frac{27}{3}\\\\h\leq 9 \ in.

Hence the height of the triangle must be at most or ≤ 9 inches.

5 0
2 years ago
Help this is really hard
scoundrel [369]

Answer:

x = 18, y = 45

Step-by-step explanation:

4x - 7 + a = 180 (linear pair)

But, a = 6x + 7 (alternate interior angles)

=> 4x - 7 + 6x + 7 = 180

=> 10x = 180

=> x = 18

Now, 3y - 20 = 6x + 7 (vertically opposite angles)

=> 3y - 20 = 6(18) + 7

=> 3y = 108 + 7 + 20

=> 3y = 135

=> y = 45

Hope it helps :)

Please mark my answer as the brainliest

4 0
2 years ago
Read 2 more answers
The weight of 3 fruit baskets are shown below. Basket A 12 Kg / Basket B 8 Kg / Basket C 18 kg Which basket is heaviest?
Feliz [49]
I think it is B I hope I am right and helped
5 0
3 years ago
To prove part of the triangle midsegment theorem using
bonufazy [111]

Answer:the length of GH is half the length of KL

Step-by-step explanation: Just took it on edg and was correct.

3 0
3 years ago
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Given ACM, angle C=90º. AP=9, PM=12. Find AC, CM, AM.
Gnesinka [82]

Answer:

AM = 25, AC = 15, CM = 20

Step-by-step explanation:

The given parameters are;

In ΔACM, ∠C = 90°, \overline{CP} ⊥ \overline{AM}, AP = 9, and PM = 16

\overline{AC}² + \overline{CM}² = \overline{AM}²

\overline{AM} = \overline{AP} + PM = 9 + 16 = 25

\overline{AM} = 25

\overline{AC}² = \overline{AP}² + \overline{CP}² = 9² +  \overline{CP}²

∴ \overline{AC}² = 9² +  \overline{CP}²

Similarly we get;

\overline{CM}² = 16² + \overline{CP}²

Therefore, we get;

\overline{AC}² + \overline{CM}² = 9² +  \overline{CP}² + 16² + \overline{CP}² = \overline{AM}² = 25²

2·\overline{CP}² = 25² - (9² + 16²) = 288

\overline{CP}² = 288/2 = 144

\overline{CP} = √144 = 12

From \overline{AC}² = 9² +  \overline{CP}², we get

\overline{AC} = √(9² +  12²) = 15

\overline{AC} = 15

From, \overline{CM}² = 16² + \overline{CP}², we get;

\overline{CM} = √(16² + 12²) = 20

\overline{CM} = 20.

3 0
2 years ago
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