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3 years ago

What the answer and how did you get it

Mathematics

1 answer:

Olegator [25]3 years ago

7 0

Let x = the cost of the student discount card.

x + the cost of the passes = 29.50

x + 15(1.50) = 29.50

x + 22.50 = 29.50

Subtract 22.50 on both sides to isolate the variable x.

x = 29.50 - 22.50

x = 7

The student discount card is $7.

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Convert 340 centimeters into inches. Round your answer to the nearest whole number.

Nana76 [90]

Answer:

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Step-by-step explanation:

6 0

3 years ago

trina planted 43 plants. she planted twice as many cucumbers as zucchini and eight more tomatoes than cucumbers. write and solve

Leto [7]

To answer this question you can create an equation in terms of the number of zucchini that were planted. Each other plant gives information that can relate to the number of zucchini plants.

# of cucumbers + # of tomatoes + # of zucchini
2z + 2z + 8 + z = 43
5z + 8 = 43
-8 -8
5z = 35
5 5
z = 7 plants

There were 7 zucchini plants, 14 (2 x 7) cucumber plants, and 22 (2 x 7 + 8) tomato plants.

5 0

3 years ago

Read 2 more answers

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 $2$ .