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3 years ago

Please help me ASAP .Picture is down below.Will mark you brainliest

Mathematics

1 answer:

Furkat [3]3 years ago

3 0

Answer:

1, 2.7

2, 3.2

3, 2.7

4, 3.2

5, 64.3

6, 64.3

Step-by-step explanation:

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Joseph built a model of a pyramid where the base of the pyramid is a square. if the base has an area of 49 cm, what is the side

Tema [17]

7, as the square root of 49 is 7

(The sides would be equal, and so the area would be one side squared)

8 0

3 years ago

PLEASE ANSWER FULL QUESTION

Komok [63]

Answer:

I don't know the answer !!! :(

8 0

3 years ago

40% of the children in a sports club play badminton. 25% of the children who play badminton also play squash. There are 11 child

Keith_Richards [23]

Answer:

There are 110 children total in the sports club

Step-by-step explanation:

To get this answer, its actually easier than it seems. You might need a calculator however.

First you start with the 11 children who play both badminton and squash, then, you divide that by 0.25 (25%) to get 44.

Next you take 44 and divide it by 0.4 (40%) to get 110.

And there you go! If you want to make sure you got it right simply start with 110 and multiply it by 0.4 then multiply the number/decimal you get by 0.25. You should get 11 to confirm your answer :)

6 0

4 years ago

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

8 0

3 years ago

Help meeeeee please chile

nasty-shy [4]

Answer:

Step-by-step explanation:

Acute angle

5 0

3 years ago

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