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3 years ago

11

MovieMania, an online movie rental website charges a one-time fee of $6.85 and $2.99 Per movie rental. Let m represent the numbe

r of movies you watch and let C represent the total cost to watch the movies. write an equation that relates the total cost to the numbers of movies you watch from MovieMania

1 answer:

nasty-shy [4]3 years ago

6 0

Answer:

C=6.85+2.99m, where:

C is the total cost to watch the movies

m is the number of movies you watch

Step-by-step explanation:

From the information provided, you can say that the total cost to watch the movies would be equal to the one-time fee plus the result of multiplying the price per movie for the number of movies you watch, which can be expressed as:

C=6.85+2.99m, where:

C is the total cost to watch the movies

m is the number of movies you watch

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Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span> $S_{a}=S_{l}-2$
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span> $t_{l}=t_{a}+1$
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span> $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

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4 years ago

How do I solve (4x/7)-(5/6)

butalik [34]

Answer:

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Step-by-step explanation:

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3 years ago

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How many nanoseconds does it take for a computer to perform one calculation if it performs 6.7 × 10^{7} calculations per second?

crimeas [40]

Answer:

<u><em>the comuter does 1 calculation in 1.5* $10^{-17}$ nanosec</em></u>

Step-by-step explanation:

<u><em>First step: we need to write all terms in nanosec, as we know:</em></u>

<u><em>1[nanosec]= $10^{-9}$ [sec]</em></u>

<u><em>The computer does 6.7* $10^{7}$ [ $\frac{calculation}{sec}$ ]</em></u>

<u><em>if we multiplicate for $\frac{sec}{nanosec*10^{*9}}$ , from the units convertion.</em></u>

<u><em>6.7* $10^{7}$ [ $\frac{calculation}{sec}$ * $\frac{sec}{nanosec*10^{*9}}$ </em></u>

<u><em>6.7* $10^{16}$ [ $\frac{calculation}{nanosec}$ ]</em></u>

<u><em>Now doing a simple three rule we have this:</em></u>

<u><em>6.7 * $10^{16}$ calculation in ⇒ 1 nanosec</em></u>

<u><em>1 calculation in ⇒ $\frac{1 nanosec* 1calculation}{6.7*10^{16} calculation }$ </em></u>

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3 years ago

Solve 3-x=-9<br><br>A. x=12<br><br>B. X= -6<br><br>C. X= 6<br><br>D. x=-12

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Answer:

D x=-12

Step-by-step explanation:

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Answer:
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Step-by-step explanation:
In a rhombus,
1 diagonal = 8 cm (left - right)
2 diagonal = 6 cm (top - bottom)

There will be 4 triangles in a rhombus when bisected with 2 intersecting diagonals (all 4 will have equal measures).

Then,
Base of 1 triangle = 8/2 = 4 cm
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Hypotenuse of 1 triangle = Longest side of triangle = Side of rhombus = ?

Using pythagorean property,
√((4)² + (3)²) = Hypotenuse
√(16 + 9) = Hypotenuse
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2 years ago

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