Answer:
I think it's 4 gigabytes
Step-by-step explanation:
He only wants to pay 65$ and he already spends 45$ on his phone already. So you only have 20 dollars until his grand totaly of 65$ so 5x4= 20 so he can only use 4 gigabytes. Does that make sense? Hope I helped eheh
Answer:
First you have to plot Bart's Garage's fee. Bart's equation is y = 30x + 20, so you would need to plot the points (0, 20) and (1, 50).
Lisa's base fee is the same, so her equation would have a + 20 at the end of it, but her hourly fee is higher. Therefore her hourly fee would be higher than 30. Her equation could be anything above 30, y = 50x +20 would work. The points for that graph are (0, 20) and (1, 70).
8h/3+19
Move all terms to the left
8-(h/3+19)=0
Get rid of parentheses
-h/3-19+8=0
Multiply all terms by denominator
-h-19*3+8*3=0
Add all numbers and variables together
-1h-33=0
Move all terms containing h to the left all other terms to the right
-h=33
h=33/-1
h=-33
The value of the tractor after 7 years is 15,000$. To find the answer, multiple .10 by 50,000 to get 5,000, then multiple 5,000 by 7 to get 35,000. Then, subtract 35,000 from 50,000 to get 15,000. Hope this helps, have a great day! Ps. Please mark me brainliest:)
It looks like we have two unknowns here and two possible equations so to solve this we will need to utilize our systems of equations, I will be using the substitution method.
We know that all of the parts multiplied together is equal to the volume.
We know that the volume is 126 square inches
The height is seven inches
The width is unknown, let w be the width
The length is unknown, let l be the length
We know that the length is three inches more than the width (l=w+3)
We know that the height times the length times the width is equal to the volume (7lw=126)
Since we know that l=w+3, we can substitute w+3 in for l,
7(w+3)(w)=126
Now we can solve our equation
7(w+3)(w)=126
(7w+21)(w)
7w^2+21w=126
Now we will complete the square to solve for w
7w^2+21w=126
w^2+3w=18
w^2+3w+2.25=20.25
(w+1.5)^2=20.25
w+1.5=4.5
w=3
We have now found out using our knowledge of systems of equations that the width of this prism is 3 inches.