16.09.2022

Write an equation for 15 is 15% of what number? Use x for your variable

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24.06.2023, solved by verified expert
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Reading through the problem, we have "15 is",

that's "15 =", 15%, 15/100, "of", times, "what number", x.

So our equation reads 15 = 15/100 · x.

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Mathematics
Step-by-step answer
P Answered by PhD

Reading through the problem, we have "15 is",

that's "15 =", 15%, 15/100, "of", times, "what number", x.

So our equation reads 15 = 15/100 · x.

Mathematics
Step-by-step answer
P Answered by PhD

y = 15x

Step-by-step explanation:

Given

Rate = 15\ miles/hour

x = time

y = distance

Required

Express in terms of y and x

The given rate is a ratio of distance to time.

So:

Ratio = Distance/Time

Substitute values for Distance and Time

Ratio = y/x

Substitute 15 for Ratio

15 = y/x

Cross Multiply

y = 15x

Hence: the situation is represented by y = 15x

Mathematics
Step-by-step answer
P Answered by PhD

y = 15x

Step-by-step explanation:

Given

Rate = 15\ miles/hour

x = time

y = distance

Required

Express in terms of y and x

The given rate is a ratio of distance to time.

So:

Ratio = Distance/Time

Substitute values for Distance and Time

Ratio = y/x

Substitute 15 for Ratio

15 = y/x

Cross Multiply

y = 15x

Hence: the situation is represented by y = 15x

Mathematics
Step-by-step answer
P Answered by PhD

Answers:

13. E 8500 < x < 12,000

14. D 6∠x∠8

15. E 1600 < x < 2200

16.  A1.02s > t > 0 s

17. 6.25 years

18. A5000 > (1500)(10)x

19. E 20.3 < 35 - 9.8t < 30.1

20. D$4167

Step-by-step explanation:

13.E 8500 < x < 12,000 ( the amount of money spent lies between 8500 and $12,000)

14. D 6∠x∠8 (subtracting 2 from both 8 and 10 will give 6 and 8 hours respectively

15. E 1600 < x < 2200 subtracting 600 calories for women

16. A1.02s > t > 0 s puttin the value of t=12,20

17. 6.25 years using the formula I=P*R*T/100

18.  A5000 > (1500)(10)x

19.E 20.3 < 35 - 9.8t < 30.1

20  D$4167

Mathematics
Step-by-step answer
P Answered by PhD
The general equation for profit is written below:

Profit = Revenue - Total Cost
Revenue is the total sales you get while total cost includes the capital and other variable costs.

Letting x be the number of t-shirts, the revenue would be 15x, while the total cost is (20 + 9x). So, the profit equation becomes:

150 ≤ 15x - (20 +9x)

The ≤ symbol means that the right side of the equation must be equal or greater than 150 so that you can reach your goal profit.
Solving for x,

150 + 20 ≤ 15x - 9x
x ≥ 28.33 
Hence, the answer is A.
Mathematics
Step-by-step answer
P Answered by PhD

Answers:

13. E 8500 < x < 12,000

14. D 6∠x∠8

15. E 1600 < x < 2200

16.  A1.02s > t > 0 s

17. 6.25 years

18. A5000 > (1500)(10)x

19. E 20.3 < 35 - 9.8t < 30.1

20. D$4167

Step-by-step explanation:

13.E 8500 < x < 12,000 ( the amount of money spent lies between 8500 and $12,000)

14. D 6∠x∠8 (subtracting 2 from both 8 and 10 will give 6 and 8 hours respectively

15. E 1600 < x < 2200 subtracting 600 calories for women

16. A1.02s > t > 0 s puttin the value of t=12,20

17. 6.25 years using the formula I=P*R*T/100

18.  A5000 > (1500)(10)x

19.E 20.3 < 35 - 9.8t < 30.1

20  D$4167

Mathematics
Step-by-step answer
P Answered by PhD
The general equation for profit is written below:

Profit = Revenue - Total Cost
Revenue is the total sales you get while total cost includes the capital and other variable costs.

Letting x be the number of t-shirts, the revenue would be 15x, while the total cost is (20 + 9x). So, the profit equation becomes:

150 ≤ 15x - (20 +9x)

The ≤ symbol means that the right side of the equation must be equal or greater than 150 so that you can reach your goal profit.
Solving for x,

150 + 20 ≤ 15x - 9x
x ≥ 28.33 
Hence, the answer is A.
Mathematics
Step-by-step answer
P Answered by PhD

x\ =\ \dfrac{209}{30}

y\ =\ \dfrac{29}{18}

z\ =\ \dfrac{64}{15}

Step-by-step explanation:

Given equations are

15x + 15y + 10z = 106

5x + 15y + 25z = 135

15x + 10y - 5z = 42

The augmented matrix by using above equations can be written as

\left[\begin{array}{ccc}15&15&10\ \ |106\\5&15&25\ \ |135\\15&10&-5|42\end{array}\right]

R_1\ \rightarrow\ \dfrac{R_1}{15}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\5&15&25|135\\15&15&-5|42\end{array}\right]

R_1\rightarrowR_2-5R1\ and\ R_3\rightarrow\ R_3-15R_1

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&10&\dfrac{65}{3}|\dfrac{299}{3}\\\\0&0&-15|-64\end{array}\right]

R_2\rightarrow\ \dfrac{R_2}{10}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&-15|-64\end{array}\right]

R_3\rightarrow\ \dfrac{R_3}{-15}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_1\rightarrow\ R_1-R_2

=\ \left[\begin{array}{ccc}1&0&\dfrac{-3}{2}|\dfrac{17}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_1\rightarrow\ R_1+\dfrac{3}{2}R_3

=\ \left[\begin{array}{ccc}1&0&0|\dfrac{209}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_2\rightarrow\ R_2-\dfrac{65}{30}R_3

=\ \left[\begin{array}{ccc}1&0&\0|\dfrac{209}{30}\\\\0&1&0|\dfrac{29}{18}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

Hence, we can write from augmented matrix,

x\ =\ \dfrac{209}{30}

y\ =\ \dfrac{29}{18}

z\ =\ \dfrac{64}{15}

Mathematics
Step-by-step answer
P Answered by PhD

see explanation

Step-by-step explanation:

Given that y varies inversely as x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition that x = 9 when y = 15

15 = \frac{k}{9} ( multiply both sides by 9 )

k = 135

y = \frac{135}{x} ← equation of variation

Mathematics
Step-by-step answer
P Answered by PhD

x\ =\ \dfrac{209}{30}

y\ =\ \dfrac{29}{18}

z\ =\ \dfrac{64}{15}

Step-by-step explanation:

Given equations are

15x + 15y + 10z = 106

5x + 15y + 25z = 135

15x + 10y - 5z = 42

The augmented matrix by using above equations can be written as

\left[\begin{array}{ccc}15&15&10\ \ |106\\5&15&25\ \ |135\\15&10&-5|42\end{array}\right]

R_1\ \rightarrow\ \dfrac{R_1}{15}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\5&15&25|135\\15&15&-5|42\end{array}\right]

R_1\rightarrowR_2-5R1\ and\ R_3\rightarrow\ R_3-15R_1

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&10&\dfrac{65}{3}|\dfrac{299}{3}\\\\0&0&-15|-64\end{array}\right]

R_2\rightarrow\ \dfrac{R_2}{10}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&-15|-64\end{array}\right]

R_3\rightarrow\ \dfrac{R_3}{-15}

=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_1\rightarrow\ R_1-R_2

=\ \left[\begin{array}{ccc}1&0&\dfrac{-3}{2}|\dfrac{17}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_1\rightarrow\ R_1+\dfrac{3}{2}R_3

=\ \left[\begin{array}{ccc}1&0&0|\dfrac{209}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

R_2\rightarrow\ R_2-\dfrac{65}{30}R_3

=\ \left[\begin{array}{ccc}1&0&\0|\dfrac{209}{30}\\\\0&1&0|\dfrac{29}{18}\\\\0&0&1|\dfrac{64}{15}\end{array}\right]

Hence, we can write from augmented matrix,

x\ =\ \dfrac{209}{30}

y\ =\ \dfrac{29}{18}

z\ =\ \dfrac{64}{15}

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