Isabel can drive 505.85 miles with a gas tank that holds 15.1 gallons of gas
<em><u>Solution:</u></em>
Given that, Isabel’s car gets 33.5 miles per gallon
Her gas tank holds 15.1 gallons of gas
To find: Number of miles can Isabel drive on a full tank of gas
Let "x" be the miles drive with 15.1 gallons of gas
From given,
1 gallon = 33.5 miles
15.1 gallon = x miles
This forms a proportion and we can solve the sum by cross multiplying
Thus Isabel can drive 505.85 miles on a full tank of gas
Answer:
Step-by-step explanation:
First,You have to realize that if you multiply the amount of food by something, you multiply the amount of calories. if some piece of food is 100 calories what happens if you eat 2? you get 2 times that number of calories. Same fi you eat half of one of that kind of food. You get half the amount of calories.
Here he makes 1 1/2 (one and a half) times the amount. Now, if this is confusing you just need to realize that multiplying by this is the same as multiplying by 1+1/2. Say, again, a piece of food is 100 calories. Multiplying it by this would look like the following.
100(1+1/2)
100*1 + 100*1/2
100+50
150
So if you ever have a mixed number like this you could split it up into an addition problem and then distribute hat you're multiplying. Another solution is to multiply by the improper fraction, which here is 3/2, so 100*3/2=150 as well. Let me know if you don't get how to get the improper fraction or how to multiply fractions.
Now, super simple, just multiply the calories by that number.
310(1+1/2) = 310(3/2)
310 + 155 = 930/2
465 = 465
Kinda showed how to multiply by fractions, still if you don't get it let me know.
Answer:
$8.82
Step-by-step explanation:
4.6 (1.15) + 0.5 (1.12) + 2.25 (1.32)
5.29 + 0.56 + 2.97
8.82
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
Now, the system of equations in matrix form will be:
![\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%26%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26%5C%5Cy%26%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%26%5C%5C7%26%5Cend%7Barray%7D%5Cright%5D%20)
Next, we are going to find the determinant:
![D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}](https://tex.z-dn.net/?f=D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%281%29%28%20%5Cfrac%7B1%7D%7B25%7D%29%20-%20%281%29%28%20%5Cfrac%7B1%7D%7B65%7D%20%29%3D%20%5Cfrac%7B8%7D%7B325%7D%20)
Next, we are going to find the determinant of x:
![D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8](https://tex.z-dn.net/?f=%20D_%7Bx%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%261%5C%5C7%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%28375%29%28%20%5Cfrac%7B1%7D%7B25%7D%20%29-%281%29%287%29%3D8)
Now, we can find x:
Now that we know the value of x, we can find y:
Remember that time equals distance over velocity; therefore, the time on the highway will be:
An the time on the city will be:
We can conclude that the bus was five hours on the highway and two hours in the city.
Answer:
x = 1/4
Step-by-step explanation:
<u>Calculations:</u>
