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3 years ago

Tell whether the system has one solution, infinitely many solutions, or no solution.

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

4 0

C. The system has no solution.

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Mary makes trail mix with bulk nuts and dried fruit. She spends $4.50 on the dried fruit. Mixed nuts cost $2.50 per pound. She h

AnnyKZ [126]

Answer: 3 pounds

Step-by-step explanation: 12 - 4.50 = 7.50 7.50/2.50=3

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3 years ago

Find the area of a regular octagon with an apothem of 8.45 cm and a side length of 7 cm. (round to the nearest whole number) A)

Alexeev081 [22]

Answer:

237

Step-by-step explanation:

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3 years ago

How many points need to be removed from this graph

ivann1987 [24]

3 is your answer mate

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3 years ago

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Compare the functions shown below: f(x) cosine graph with points at 0, negative 1 and pi over 2, 1 and pi, 3 and 3 pi over 2, 1

Levart [38]

Answer:

f(x) and h(x) have the same maximum value: 3

Step-by-step explanation:

The maximum value of f(x) is 3 at (π, 3).

The maximum value of g(x) is -2 at (-3, -2).

The maximum value of h(x) is 3 at (0, 3).

Both f(x) and h(x) have the same (greatest) maximum value.

7 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