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2 years ago

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The total cost for buying new sneakers for the basketball team an be found by using the function y=25x + 65, is where x is the n

umber of players. If between 9 and 16 players need new sneakers. Find the domain and range of this situation

1 answer:

gtnhenbr [62]2 years ago

3 0

Answer:

$Domain\ =\ \{x: 9,10,11....16\}$

<em></em> $Range\ =\ \{y:290, 315,340.....465\}$ <em></em>

Step-by-step explanation:

Given

$y = 25x + 65$

Required

Determine the range and domain when x = 9 to 16

The domain is the set of

$Domain\ =\ \{x: 9,10,11....16\}$

Solving for the range;

<em>When x = 9</em>

<em></em> $y = 25 * 9 + 65$ <em></em>

<em></em> $y = 225 + 65$ <em></em>

<em></em> $y = 290$ <em></em>

<em></em>

<em>When x = 10</em>

<em></em> $y = 25 * 10 + 65$ <em></em>

<em></em> $y = 250 + 65$ <em></em>

<em></em> $y = 315$ <em></em>

<em></em>

<em>When x = 11</em>

<em></em> $y = 25 * 11 + 65$ <em></em>

<em></em> $y = 275 + 65$ <em></em>

<em></em> $y = 340$ <em></em>

<em></em>

<em>You continue till you get to x = 16</em>

<em>---</em>

<em>--</em>

<em>-</em>

<em>When x = 16</em>

<em></em> $y = 25 * 16 + 65$ <em></em>

<em></em> $y = 400 + 65$ <em></em>

<em></em> $y = 465$ <em></em>

<em></em>

<em>Hence; the range is</em>

<em></em> $Range\ =\ \{y:290, 315,340.....465\}$ <em></em>

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A caterer makes 15 3/4 quarts of tomato soup. He pours 2 cups of soup into each bowl. How many quarts of soup are in each bowl?

Sindrei [870]

<u>ANSWER: </u>

There are 0.5 quarts in each bowl. There are 31 fully filled bowls. And left over soup is 1 cup of soup.

<u>SOLUTION: </u>

Given, a caterer makes $15 \frac{3}{4}$ which is equal to $\frac{63}{4}$ quarts of tomato soup.

He pours 2 cups of soup into each bowl.

Now let us find how many quarts of soup are in each bowl?

We know that, 1 us liquid quart equals to 4 us cups.

Then, 1 quart = 4 cups

1 cup = $\frac{1}{4}$ quarts

Here, in each bowl caterer pours two cups,

So, 2 cups = $2 \times \frac{1}{4}$ quarts

$= \frac{1}{2} = 0.5 quarts$

Hence, there are 0.5 quarts in each bowl.

Now let us find how many 2-cup bowls of soup can the caterer fill to the top?

From above part, we know that, each bowl has 0.5 quarts.

Then, total bowls with 2- cups of soup = $\frac{\text {total soup}}{\text {soup in one bowl}}$

$=\frac{\frac{63}{4}}{\frac{1}{2}}$

$=\frac{63}{4} \times \frac{2}{1}$$=\frac{63}{2}$$=31.5$$

Here, 0.5 means that a bowl which is not filled to the top. So there are 31 fully filled bowls.

Now, let us find how many cups of soup are left over?

From above part we saw that, there are 31 fully filled bowls but there is a bowl with 0.5 quarts and which is not fully filled.

So, it is the amount of soup which is left over. i.e. 0.5 times of a bowl.

We know that, each bowl will have two cups.

Here left over bowl has half of it filled with soup

So the left over soup is 1 cup of soup.

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2 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

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Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

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