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3 years ago

5

rogelio has car insurance that he must pay four times a year. If each payment is $144, how much money should he set aside each m

onth to cover his car insurance?

2 answers:

RSB [31]3 years ago

7 0

He should set aside at least $48 each month.

Neko [114]3 years ago

5 0

Answer: 48$
Explanation: since he pays his car insurance 4 times a year and each payment is 144$ that means he pays 144$ every 3 months since 12 divided by 4 equals 3
So 144 divided by 3 is 48 so if he puts aside 48 dollars each month he will have 144$ every 3 month to pay his car insurance

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What is 8 21/40 as a fraction or mixed number as a decimal

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3 years ago

On her lunch break, Audrey grabs a meal with her two coworkers. If everyone paid an equal amount of $8.50, then how much was the

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Answer:

$25.50

Step-by-step explanation:

3 people including Audrey went on the lunch break. If each spent $8.50, the total bill can be calculated in two ways

<u>Method 1 </u>

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<u>Method 2 </u>

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Step-by-step explanation:

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3 years ago

the area of a triangle is 6 and 3/8 square yards. The height of the triangle is 2 and 1/2 yards. What is the length of the base

Otrada [13]

Solution:

<u>Note that:</u>

Area of triangle = 6 3/8 yd²
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<u>Use the formula to find the base of the triangle.</u>

1/2 x base x altitude = 6 3/8 yd²
=> 1/2 x base x 2 1/2 = 6 3/8 yd²
=> 1/2 x base x 5/2 = 6 3/8 yd²
=> 5/4 x base = 6 3/8 yd²
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2 years ago

Read 2 more answers

Find the point P on the graph of the function y=√x closest to the point (9,0)

Sphinxa [80]

Answer:

$\displaystyle \frac{17}{2}$ .

Step-by-step explanation:

Let the $x$ -coordinate of $P$ be $t$ . For $P\!$ to be on the graph of the function $y = \sqrt{x}$ , the $y$ -coordinate of $\! P$ would need to be $\sqrt{t}$ . Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$ .

The Euclidean Distance between $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ is:

$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$ .

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$ . Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$ , would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ .

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$ :

$\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17$ .

$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$ .

Set the first derivative, $(2\, t - 17)$ , to $0$ and solve for $t$ :

$2\, t - 17 = 0$ .

$\displaystyle t = \frac{17}{2}$ .

Notice that the second derivative is greater than $0$ for this $t$ . Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$ . This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ , the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ .

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$ -coordinate of $P\!$ is $\displaystyle \frac{17}{2}$ .

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3 years ago