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Artyom0805 [142]
2 years ago
12

Help me solve this problem please

Mathematics
1 answer:
cricket20 [7]2 years ago
4 0

Answer:

D. 36.75-3x

Step-by-step explanation:

You have a total amount of 36.75.

Since you are spending money you need to subtract.

We know that Marshall buys 3 tickets.

We also know that 'x represents the amount of a single ticket.

Remember that they are asking you for an algebraic expression and not an equation.

So you need to take away from your total, 36.75. And since you plan on buying three tickets and aren't given the amount x it leaves you with taking 3x from your total 36.75 or an other words

D. 36.75 - 3x

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Answer:

Y=4/3x-4

Step-by-step explanation:

You have it in slope intercept form =)

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A ladder reaches 2m up a wall. The foot of ladder is 0.4m from the wall. For safety reasons, the slope should be between 6.3 and
NeTakaya

Answer:

The slope of ladder is not within the safe range  .

Step-by-step explanation:

Given as :

The height of the ladder up a wall = h = 2 meters

The distance of base of ladder from the wall = x = 0.4 meters

The estimated safety range of slope of ladder is between 6.3 and 9.5

Let The slope of the ladder = x

<u>According to question</u>

slope of the ladder = \dfrac{\textrm height of ladder up wall}{\textrm distance of base of ladder from wall}

Or, x = \dfrac{2 meters}{0.4 meters}

Or, x = 5

So, The slope of the ladder = x = 5

Since The estimated range of slope between 6.3 and 9.5

And The calculate slope of ladder = 5

Hence, The slope of ladder is not within the safe range  . Answer

4 0
4 years ago
A loan of $1,020 was repaid at the end of 18 months with a check for $1,050. What annual rate of interest was charged?
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We'll be assuming this is with simple interest.

now, a year has 12 months, thus in 18 months there are 18/12 years.

\bf ~~~~~~ \textit{Simple Interest Earned Amount}&#10;\\\\&#10;A=P(1+rt)\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$1050\\&#10;P=\textit{original amount deposited}\to& \$1020\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=years\to \frac{18}{12}\to &\frac{3}{2}&#10;\end{cases}

\bf 1050=1020\left[ 1+r\left( \frac{3}{2} \right) \right]\implies \cfrac{1050}{1020}=1+r\left( \frac{3}{2} \right) \implies \cfrac{35}{34}=1+r\left( \frac{3}{2} \right)&#10;\\\\\\&#10;\cfrac{35}{34}-1=\cfrac{3}{2}r\implies \cfrac{1}{34}=\cfrac{3}{2}r\implies \cfrac{2\cdot 1}{3\cdot 34}=r\implies \cfrac{1}{51}=r&#10;\\\\\\&#10;r\%=100\cdot \cfrac{1}{51}\implies r\approx\stackrel{\%}{1.96078431372549}
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What is an appropriate conclusion to draw from the following statements:
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<span>A. Sally, Bob, and Javier ride the bus every day.</span>
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