Csn someone please help?

marta [7]

1A
2D
3B
4C
5E

Hope this helps!

3 0

3 years ago

A family has two cars during one particular week the first car consume 25 gallons of gas and the second consumed 15 gallons of g

d1i1m1o1n [39]

Answer: the efficiency of the first car is 25 miles per gallon.

the efficiency of the second car is 30 miles per gallon.

Step-by-step explanation:

Let x represent the efficiency of the first car.

Let y represent the efficiency of the second car.

Distance = car efficiency × number of gallons.

The first car consume 25 gallons of gas and the second consumed 15 gallons of gas. The two cars Drove a combined total of 1075 miles. It means that

25x + 15y = 1075- - - - - - - - - - -1

The sum of the fuel efficiencies was 55 miles per gallon. It means that

x + y = 55

Substituting x = 55 - y into equation 1, it becomes

25(55 - y) + 15y = 1075

1375 - 25y + 15y = 1075

- 25y + 15y = 1075 - 1375

- 10y = - 300

y = - 300/-10

y = 30

x = 55 - y = 55 - 30

x = 25

4 0

3 years ago

Six times a number is greater than 20 more than that number. Which are the possible values of that number?

Neporo4naja [7]

I don't know how to make a 'greater than' sign.

6x is greater than x + 20
6x - x is greater than 20
5x is greater than 20
x is greater than 20 divided by 5
x is greater than 4

So, any number greater than 4 will work in this equation.

4 0

3 years ago

Read 2 more answers

Roman is born in the year 123 BC and dies at the age of 74 years. Use a negative number to express the year in which he dies

Stells [14]

Answer:

-49 or 49 BC

hope that this helps

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