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4 years ago

Jane’s favorite fruit punch consists of pear, pineapple, and plum juices in the ratio 5:2:3.

1 answer:

3 0

5:2:3
5+2+3=10

If she has double of plum juice the ratio will be

10:4:6
10+4+6=20

So she can make 20 cups of fruit punch.

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A.Find the total and area of the compound figures

goldfiish [28.3K]

Answer:

(a) The total area of compound figure is $284.52ft^2$ .

(b) The total perimeter of compound figure is 68.84 ft.

Step-by-step explanation:

Perimeter: The total distance covered by the figure.

Area: It is defined as the space occupied by the figure.

Given:

Length of rectangular figure = 19 ft

Breadth of rectangular figure = 12 ft

Radius of circle, r = $\frac{12ft}{2} = 6 ft$

Part (a):

Now calculating the total area of the compound figure.

$\text{Total area}=\text{Area of rectangular figure}+\frac{1}{2}\times \text{Area of circle}\\\\\text{Total area}=\text{Length}\times \text{Breadth}+\frac{1}{2}\times \pi r^2$

Now putting all the given values in this formula, we get:

$\text{Total area}=(19ft\times 12ft)+\frac{1}{2}\times 3.14\times (6ft)^2\\\\\text{Total area}=284.52ft^2$

Part (b):

Now calculating the total perimeter of the compound figure.

$\text{Total perimeter}=2\times \text{Length}+Breadth+(\frac{1}{2}\times \text{Circumference})\\\\\text{Total perimeter}=2\times \text{Length}+Breadth+(\frac{1}{2}\times 2\times \pi \times r)$

Now putting all the given values in this formula, we get:

$\text{Total perimeter}=2\times \text{Length}+Breadth+(\frac{1}{2}\times 2\times \pi \times r)\\\\\text{Total perimeter}=2\times 19ft+12ft+(\frac{1}{2}\times 2\times 3.14 \times 6ft)\\\\\text{Total perimeter}=68.84ft$

8 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

PLEASE HELP!!!<br> Solve the inequality: <br> 5 (2x-3) - 6x > -2 (x - 6) +3

jeyben [28]

Answer:

x>5

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

4x−15>−2x+15

Step 2: Add 2x to both sides.

4x−15+2x>−2x+15+2x

6x−15>15