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lozanna [386]
2 years ago
9

Kevin and his children went into a restaurant and he bought $31.50 worth of hotdogs and drinks. Each hotdog costs $4.50 and each

drink costs $2.25. He bought 3 times as many hotdogs as drinks. Graphically solve a system of equations in order to determine the number of hotdogs, x,x, and the number of drinks, y,y, that Kevin bought.
Mathematics
1 answer:
Musya8 [376]2 years ago
6 0

Answer:

6 hot dogs, 2 drinks

Step-by-step explanation:

let y = number of hot dogs

let x = number of drinks

system:

y = 3x

4.5y + 2.25x = 31.50   (standard form:  y = -1/2x + 7)

if you graph the system, lines intersect at (2, 6)

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Principle = 30,000<br>Time = 4 years<br>Rate = 30<br>Then find the simple interest ?​
Nutka1998 [239]

Answer:

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Given:}}}}}}\end{gathered}

  • {\dashrightarrow \sf{Principle = Rs.30000}}
  • {\dashrightarrow \sf {Time = 4 \: years}}
  • \dashrightarrow \sf{Rate = 30\%}
  • \begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{To Find:}}}}}}\end{gathered}

  • \dashrightarrow{\sf{Simple \: Interest }}

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Using Formula:}}}}}}\end{gathered}

\dag{\underline{\boxed{\sf{ S.I = \dfrac{P \times R \times T}{100}}}}}

Where

  • \dashrightarrow{\sf{S.I = Simple \:  Interest }}
  • {\dashrightarrow{\sf{P = Principle }}}
  • {\dashrightarrow{\sf{ R = Rate }}}
  • {\dashrightarrow{\sf{T = Time}}}

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Solution:}}}}}}\end{gathered}

{\quad {: \implies{\sf{ S.I =  \bf{\dfrac{P \times R \times T}{100}}}}}}

Substituting the values

{\quad {: \implies{\sf{ S.I =  \bf{\dfrac{30000 \times 30\times 4}{100}}}}}}

{\quad {: \implies{\sf{ S.I =  \bf{\dfrac{30000 \times 120}{100}}}}}}

{\quad {: \implies{\sf{ S.I =  \bf{\dfrac{3600000}{100}}}}}}

{\quad {: \implies{\sf{ S.I =  \bf{\cancel{\dfrac{3600000}{100}}}}}}}

{\quad {: \implies{\sf{ S.I =  \bf{Rs.36000}}}}}

{\dag{\underline{\boxed{\sf{ S.I ={Rs.36000}}}}}}

  • Henceforth,The Simple Interest is Rs.36000..

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline\color{brown}{Learn More:}}}}}}\end{gathered}

\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \dag \: \underline{\bf{More \: Useful \: Formula}}\\ {\boxed{\begin{array}{cc}\dashrightarrow {\sf{Amount = Principle + Interest}} \\ \\ \dashrightarrow \sf{ P=Amount - Interest }\\ \\ \dashrightarrow \sf{ S.I = \dfrac{P \times R \times T}{100}} \\ \\ \dashrightarrow \sf{P = \dfrac{Interest \times 100 }{Time \times Rate}} \\ \\ \dashrightarrow \sf{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}} \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

8 0
3 years ago
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Question is in the picture.
motikmotik

I think the perimeter is 64. If you divided the 4 sides of the window by 256, you get 64. Sorry if it's not right. I hope it is. c;

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3 years ago
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The population of a colony of bacteria can be modeled by the function P(t) = 19,300(5)^t, where t is the elapsed time
allsm [11]
Answer: root t(5)/19300

Explanation:

P(t) = 19300(5)^t
P-1(t) = root t(5)/19300


4 0
3 years ago
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Suppose yoko borrows $3500 at an interest rate of 4% compounded each year Find the amount owed at the end of 1 year
olchik [2.2K]

Amount owed at the end of 1 year is 3640

<h3><u>Solution:</u></h3>

Given that yoko borrows $3500.

Rate of interest charged is 4% compounded each year

Need to determine amount owed at the end of 1 year.

In our case :

Borrowed Amount that is principal P = $3500

Rate of interest r = 4%

Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1

<em><u>Formula for Amount of compounded yearly is as follows:</u></em>

A=p\left(1+\frac{r}{100}\right)^{n}

Where "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula we get

\mathrm{A}=3500\left(1+\frac{4}{100}\right)^{1}

\begin{array}{l}{A=\frac{3500 \times 104}{100}} \\\\ {A=35 \times 104} \\\\ {A=3640}\end{array}

Hence amount owed at the end of 1 year is 3640

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3 years ago
Please help with both questions, thank you!
algol [13]

11) 3 (2x + 7) - 4 = 53 - 3x is the given

Then, we use distributive property and get:

6x + 21 - 4 = 53 - 3x

Next, we combine like terms

6x + 17 = 53 - 3x

Then, we use the addition property of equality to add 3x to both sides.

9x + 17 = 53

We then use the subtraction property of equality to subtract 17 from both sides.

9x = 36

Finally, we use the division property of equality to divide 36 by 9, which isolates x.

x = 4

So, (In order) you have the: Given, Distributive Property, Combine Like Terms, Addition Property of Equality, Subtraction Property of Equality, and Division Property of Equality.

12) Please Note: This is an inequality problem. Which means we do not use the properties of equality. We'd use the properties of order. Properties of order are specific to inequalities, as the properties of equality are specific to the equations. They do not cross. As for your paper, I'm not sure why it says to use the properties of equality. I will be using the properties of order.

-7 (3 + 8x) + 5x ≤ 1 - 62x is the given

-21 - 56x + 5x ≤ 1 - 62x is distributive property

We then combine like terms

-21 - 51x ≤ 1 - 62x

Next, we use subtraction property of order to subtract 1 from each side.

-22 - 51x ≤ -62x

Then, we add 51x to each siding using the addition property of order

-22 ≤ -11

Lastly, we use the division property of order to divide 11 from both sides, which then gives us:

2 ≥ x

It becomes a positive because we're dividing 2 negatives, and we also turn the sign because we're dividing by negatives.

In order: Given, Distributive Property, Combine Like Terms, Subtraction Property of Order, Addition Property of Order, and Division Property of Order.

 

5 0
3 years ago
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