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CaHeK987 [17]
3 years ago
8

4. The admission fee at a small fair is $1.50 for

Mathematics
1 answer:
inysia [295]3 years ago
7 0

Answer:

The answer is a

Step-by-step explanation:

500 times 4 is 2,000

986 times 1.50 is 1,479

Therefore your answer is a

¯\(°_o)/¯

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HELPP 10 POINTS Here's a graph of a linear function. Write the equation that describes that function.
erastova [34]

m = (y2-y1)/(x2-x1)

m = (6-4)/(4-0)

m = 2/4

y = mx + c

6 = (2/4)(4) + c

c = 4

Thus, the equation is y = (2/4)x +4

3 0
2 years ago
Amos bought 5 cantaloupes for $8. How many cantaloupes can he buy for $24?
olchik [2.2K]
Amos can buy 15 cantaloupes
5 0
2 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.

Apparently \displaystyle c = \frac{16}{10} = \frac{8}{5} works. c = 1.6 while \displaystyle \frac{1}{c} = \frac{5}{8} = 0.625.

8 0
3 years ago
Read 2 more answers
In A QRS, ZRTQ - STR.<br> R<br> S<br> Which relationship proves that ZQRS is a right angle?
Reil [10]

Answer:

A

Step-by-step explanation:

hhjjjjkkkdaa vhjkkjddgj hjkjcgh

4 0
2 years ago
The Capulet and Montague families love writing. Last year, each Capulet wrote 4 essays, each Montague wrote 6 essays, and both f
Eduardwww [97]

Answer:

Infinite solution

Step-by-step explanation:

Let, number of Capulet family = x and number of Montague family = y.

Since, last year, each Capulet wrote 4 essays and each Montague wrote 6 essays.

Moreover, total essays wrote last year are 100.

So, we get, 4x + 6y = 100.

Again, this year, each Capulet wrote 8 essays and each Montague wrote 12 essays.

Moreover, total essays wrote this year are 200.

So, we get, 8x + 12y = 200.

Thus, the system of equations is given by,

4x + 6y = 100.

8x + 12y = 200.

Dividing first equation by 2 and second equation by 4, we get the equation,

2x + 3y = 50.

Since, there is only one equation and two variables i.e. x and y.

There can be infinite number of possibilities for the values of x and y.

6 0
3 years ago
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