Answer:
Perimeter of the picture and frame = 38.4inches
Area of the picture and frame = 92.16inches²
Step-by-step explanation:
A square frame is made up of 4 different pieces. The shape of each piece = Rectangle
The perimeter of the rectangle = 24
Perimeter of the rectangle = 24 inches
The perimeter of a rectangle = 2L + 2W
The Width of a Rectangle is always on her than the length hence.
24 = 2L + 2W
24 = 2( L + W)
24/2 = L + W
12 = L + W
Because the width is always longer than the length
W > L
Width of wooden frame = 4 × Length
Therefore;
4 × L = W
Which gives
L + W = 12 inches
4 × L + L = 12 inches
L×(4 + 1)
= 5L = 12 inches
L = 12/5 = 2.4 inches
W = 4 × L = 4 × 12/5
W = 48/5 = 9.6 inches
Side length of wooden frame, L =9.6
The perimeter of the picture frame = 4 × L= 4 × 9.6= 38.4 inches
The area of the picture frame = L²
= L × L
= 9.6 × 9.6 = 92.16inches².
Answer:
A) 277 million years
Step-by-step explanation:
Here, decay of U-238 is first order reaction. (since most of the nuclear decays are first order).
so, the equation t = × ln()
where, t = time from start of reaction
k = reaction constant
A(0) = initial concentration; A = present concentration.
⇒ 4.5 billion = 4500 million years = × ln2.
Now, given 95.82% of parent U-238 isotope is left.
⇒ t = × ln()
t = × ln()
⇒ t ≅ 277 million years.
Answer:
x = -3
Step-by-step explanation:
12 -4x-5x = 39
Combine like terms
12 - 9x = 39
Subtract 12 from each side
12-9x-12 = 39-12
-9x = 27
Divide by -9
-9x/-9 = 27/-9
x = -3
Answer:
523.6 cm^3
Step-by-step explanation:
We know the diameter is 10. The radius is half the diameter, so 10/2=5.
Knowing the radius is 5, we just need to plug it into the formula.
Volume=(4/3)pi5^3
≈523.6
Answer:523.6 cm^3
Information Given: She has 12 groups of sheep, 3 groups of cows and 2 groups of pigs. Question asks: There are 150 more sheep than pigs. How many cows has she got?
They figured 150 more sheep than pigs from 2 pigs X 75 sheep=150 lambs of 12 groups of sheep.
150 lambs/3 groups of cows=50 cows.
She has 50 cows from 3 groups of cows.