Answer:
1/12 of the container of milk will be left in the container
Step-by-step explanation:
To solve the question, we first note the variables
Volume of milk available = 9/12 of a container
Volume of milk poured in a pan = 2/3 of a container
Quantity of milk left = Initial quantity of milk less the quantity poured out
That is 9/12-2/3 = 1/12 of the container of milk
This means that there will be only 1/12 of the container of milk left
Answer:
-x + 1 > 3, and 4x + 6 < -2
Step-by-step explanation:
The inequality reads x < -2, so the arrow follows any values that are less than -2. So we need to simplify each inequality in the following options to see which match x < -2.
-4x+3> -5 simplifies to x < 2 so it's wrong
-2x + 4 > -4 simplifies to x < 4 so it's wrong
-x + 1 > 3 simplifies to x < -2, so it's right
2x + 5 < -3 simplifies to x < -4, so it's wrong
4x+ 6 < -2 simplifies to x < -2, so it's right
It’s a wide-ranging, completely unfounded theory that says that president trump is waging a secret war against elite satan-worshipping in paeophiles in government, business and the media.
Answer/Step-by-step explanation:
1. The figure is composed of a triangle and a rectangle.
Area of the triangle = ½*base*height
base = 4 ft
height = 12 - 8 = 4ft
Area of triangle = ½*4*4 = 8 ft²
Area of rectangle = length * width
Length = 8 ft
Width = 4 ft
Area of rectangle = 8*4 = 32 ft²
✔️Area of the figure = 8 + 32 = 40 ft²
2. The figure is composed of a semicircle and a triangle
Area of the semicircle = ½(πr²)
radius (r) = 3 cm
π = 3
Area = ½(3*3²) = 13.5 cm²
Area of triangle = ½*base*height
base = 3*2 = 6 cm
height = 6 cm
Area = ½*6*6 = 6 cm²
✔️Area of the figure = 13.5 + 6 = 19.5 cm²
Hi,
x + y = 350 (equation 1)
3x + 2y = 950 (equation 2)
Use Substitution Method
Isolate variable x in equation 1
x = 350 – y (equation 3)
Substitute equation 3 into equation 2
3(350 – y) + 2y = 950
1050 – 3y + 2y = 950
3y – 2y = 1050 – 950
y = 100
Substitute y = 100 into equation 1
x + 100 = 350
x = 250
Answer: 250 $3 tickets and 100 $2 tickets were sold.
