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3 years ago

If one liter of paint covers 100 square feet, how many gallons of paint will cover 800 square meters?

1 answer:

6 0

It takes 8 liters to cover 800 sqft of floor

there are 3.78541 liters in a gallon so it would take 2.11338 gallons to cover the floor

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Please help me i need an answer

kow [346]

Julie wants to buy cartons of apple juice with $55. Each carton costs $3.75.
Let x be the number of cartons she can buy.

3.75x =< 55
Divide both sides by 3.75
3.75x / 3.75 =< 55/3.75
x =< 14.66666

Of course, she can't buy 14.666 cartons so we can the integer part only. Julie can buy 14 cartons of Juice.

Hope this helps! :D

8 0

3 years ago

HELP ASAPPPP<br><br><br> please help me

Flauer [41]

Answer:

x=3, y=4

Step-by-step explanation:

hope this helps!

pls mark brainliest

4 0

2 years ago

Exponet for 13 ·13·13·13

photoshop1234 [79]

The exponent for that would be 13^3.

If the answer is incorrect, then i am sorry.

6 0

3 years ago

Read 2 more answers

A parking lot has a total of 60 cars and trucks. the ratio of cars to trucks is 7:3. how many cars are in the parking lot? how m

Vinil7 [7]

A) c + t = 60
B) 3c = 7t
Multiplying A) by -3
A) -3c + -3t = -180 then adding this to b
B) 3c = 7t
180 = 10t
trucks = 18
cars = 42

7 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago