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3 years ago

10x10-5(300x2)/5 Ian theb jkkkkj

Mathematics

1 answer:

Len [333]3 years ago

4 0

Answer:

=10x¹⁰−300x²

Step-by-step explanation:

Refer to the Attachment for steps.

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Jk=8x+6 and kL=6x+20 Find JL

mr Goodwill [35]

Jk=8x+6
J=(8x+6)/k

kL=6x+20
L=(6x+20)/k

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4 years ago

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3 years ago

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DanielleElmas [232]

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3 years ago

Prestige Builders has a development of new homes. There are four different floor plans, seven exterior colors, and an option of

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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]

Question:

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?

Answer:

36 child tickets were sold

Solution:

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

3 years ago