Question 5: A car travel a distance of 330km at an average speed of to

tresset_1 [31]

3 possible solutions to the equation 2x+y=10 are shown below:

X= 3, Y=4
(3*2+4=10)

X= 1, Y=8
(1*2+8=10)

X=4, Y=2
(4*2+2=10)

Hope this helps! :)

3 0

3 years ago

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-4(5k-7)-2k=-2(9k+5)

Mice21 [21]

If you distribute -4 through (5k-7), it equals -20k + 28. Then put on the other -2k on that side of the equals sign and you should have -22k + 28 = -2(9k+5) Then you need to distribute -2 through (9k+5) which equals -18k - 10. You then should have -22k+28=-18k-10. Add 22k to both sides to cancel it out, then you should have 28=4k-10. Add 10 to both sides to get the 4k by itself, which will be 38=4k, then divide both sides by 4 to get k by itself. 38/4 = 19/2. k = 19/2

5 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

Explain how to find the area of an enlarged polygon if you know the area of the original polygon and the scale factor of the enl

julia-pushkina [17]

Yo just need to multiply area to that scale factor,

5 0

3 years ago

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A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the

allochka39001 [22]

Answer:

The receiver must be placed at a distance of 6.125 ft from the center.

Step-by-step explanation:

The dish is of parabola shape with the length of 14 feet and 2 feet deep from the center.

A parabola is a curve which is a set of all points in the plane which are at equal distance away from a given point called focus and given line called directrix.

We have to find out the point at which all the waves comes to one point where the receiver must be placed is the focus of the parabola.

The above scenario is shown in the image which can be converted in mathematical graphical representation of the parabolic curve.

The extreme two end points becomes, (-7,2) and (7,2)

The general equation for such parabola is:

x² = 4ay

where, a is the distance from the center and the focus.

Since, (7,2) satisfy the equation so,

7² = 4a×2

a = 49 /8

Thus,

<u>The receiver must be placed at a distance of 6.125 ft from the center.</u>

5 0

2 years ago