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romanna [79]
4 years ago
8

HELP ME WITH MY LAST PROBLEM PLEASE!!!! I REALLY NEED HELP!! I DON"T KNOW HOW TO DO MY LAST PROBLEM!!!! PLEASE HELP ME!!!

Mathematics
1 answer:
kompoz [17]4 years ago
7 0
Dude wheres the question, you have to ask legistic questions this is a school website not a circus

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I’m so confused. I need help :C
svlad2 [7]

Answer:

Apple A

Step-by-step explanation:

Apple A 6.90/6 = <u>1.15 lb</u>

Apple B 12oz/ (16oz or 1lb) =3/4   .90/(3/4) =<u>1.2lb</u>

Apple C 0.45/(1/4)= <u>1.8lb</u>

3 0
3 years ago
2) Find the area and perimeter of a semi -circular disc whose radius is 10cm. (take π = 3.14)
SVETLANKA909090 [29]

Answer:

<h3>Area =Pi r2</h3>

=3.14×10x10

=3.14×100cm2

=3.14cm2

<h3>Perimeter = 2 pi r</h3>

=2 ×3.14×10

=6.28cm

Or =Pi d

= 3.14 ×20cm

=62.8cm

<h3>Note:</h3>

R means radius

D means diameter

radius is half of a diameter e.g radius is10 and diameter is 20

I can recommend you watching yt videos it helped me alot

hoo this helps:)

5 0
3 years ago
Read 2 more answers
At the city museumy child admission is and admission is $9.30. On Monday four times as many adult tickets as child tickets were
Scilla [17]

<em><u>Question:</u></em>

At the city museum, child admission is $5.80 and adult admission is $9.30. On Monday, four times as many adult tickets as child tickets were sold, for a total sales of $1548.00. How many child tickets were sold that day?

<em><u>Answer:</u></em>

36 child tickets were sold

<em><u>Solution:</u></em>

Given that,

Cost of 1 child admission = $ 5.80

Cost of 1 adult admission = $ 9.30

Let "c" be the number of child tickets sold

Let "a" be the number of adult tickets sold

On Monday, four times as many adult tickets as child tickets were sold

Number of adult tickets sold = four times the number of child tickets

Number of adult tickets sold = 4(number of child tickets sold)

a = 4c ----- eq 1

They were sold for a total sales of $ 1548.00

number of child tickets sold x Cost of 1 child admission + number of adult tickets sold x Cost of 1 adult admission = 1548.00

c \times 5.80 + a \times 9.30 = 1548

5.8c + 9.3a = 1548  ---- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "c" and "a"

Substitute eqn 1 in eqn 2

5.8c + 9.3(4c) = 1548

5.8c + 37.2c = 1548

43c = 1548

c = 36

Thus 36 child tickets were sold that day

6 0
4 years ago
What are the lengths of the legs of a right triangle in which one acute angle measures 19o and the hypotenuse is 15 units long?
WITCHER [35]
x,y-\ lengths\ of\ legs\\\\sin19^\circ=\frac{x}{15}\\\\ sin19^\circ*15=x\\\\0,32*15=x\\\\x=4,8\\\\&#10;y^2+(4,8)^2=15^2\\\\y^2=225-23,04\\\\ y^2=201,96\\\\y=14,2\\\\Legs\ are\ equalt\ to:\ 4,8 \ and\ 14,2.
6 0
4 years ago
Find the length of c for both. Please help!
hammer [34]
The answer is c = 15
4 0
3 years ago
Read 2 more answers
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