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3 years ago

7

Jake paid $13.50 for admission to the county fair and bought 9 tickets to play games. If he spent a total of $36, what is the co

st, c, of one ticket? Write and solve an equation.

2 answers:

Kipish [7]3 years ago

6 0

Answer:

the answer is 5.50

Step-by-step explanation:

hsbsbyehs

r-ruslan [8.4K]3 years ago

4 0

Answer:

2.50

Step-by-step explanation:

36 - 13.50 = 22.50

then divide that by 9 = 2.50

You might be interested in

aleksklad [387]

Answer:

4.12 kg

Step-by-step explanation:

Regular cakes:

1 dozen normal sponge cakes: 264 g plain flour

Vegetarian cakes:

1 dozen cakes: 264 g plain flour

4 eggs are replaced by 4 * 30 g of flour = 120 g flour

total flour for 1 dozen vegetarian cakes = 264 g + 120 g = 384 g

Proportion for regular cakes:

12 cakes to 264 g flour = 100 cakes to x grams flour

12/264 = 100/x

12x = 26400

x = 2200

2200 g flour for 100 regular cakes

Proportion for vegetarian cakes:

12 cakes to 384 g flour = 60 cakes to y grams flour

12/384 = 60/y

12y = 384 * 60

12y = 23040

y = 1920

1920 g flour for 60 vegetarian cakes

Total flour needed:

2200 g + 1920 g = 4120 g

4120 g * 1 kg/(1000 g) = 4.12 kg

Answer: 4.12 kg

4 0

3 years ago

Divide the given polynomial by the given monomial 3 x square - 2 x divided by X

inna [77]

Dividing (3x^2-2x) by x gives out 3x-2

7 0

3 years ago

The histogram below shows the number of weeks Jimmy spent different amounts of money to fill the gas tank in his car. If the gra

sammy [17]

Answer:

A. The difference in the heights of the bars will not appear as great.

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago

Mrs.Oliveros bought 12.036 gallons of gasoline for her car.How is this number written in expaned notation

iren [92.7K]

10+2+0.0+0.03+0.006=12.036

4 0

3 years ago