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3 years ago

There are 60 inches in 5feet. How many inches are in 8 feet ?

Mathematics

1 answer:

Murljashka [212]3 years ago

7 0

Answer:

Step-by-step explanation:

Feet have 12 inches each to find the amount in x amount of inches let this equation help you

X=amount of inches

y= amount of feet

X= 12·y

in other word multiply 12 by the amount of feet to get your answer

good luck :)

also... Brainliest?

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Katie is going to a carnival and received $50 from her mother to spend. At the carnival, entry costs $12.00 and ride tickets cos

nlexa [21]

Answer: See explanation

Step-by-step explanation:

Your question isn't complete but I believe that you want to know the number of rides that Katie can take.

Based on the information in the question, we can form an equation. Let the number of rides that Katie can take be represented by x. Therefore,

12 + (1.25 × x) = 50

12 + 1.25x = 50

1.25x = 50 - 12

1.25x = 38

x = 38/1.25

x = 30.4

Therefore, Katie can take at most 30 rides.

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3 years ago

- 2(w+6) +9< - 15<br><br>I'm make you brainliest

navik [9.2K]

It is 9lt-2w-27! Hope it helped!

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3 years ago

Read 2 more answers

Need an answer to both please

ANTONII [103]

Plz mark me the braniliest

Answer:

1st answer: 52 serves

2nd answer: $8.40

Step-by-step explanation:

1st: 80 times 65 equals 5,200 divided by 100 equals 52

2nd: 42 times 0.20 equals 8.4 or 8.40

6 0

3 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .