Answer:
3 m : 4 cm, 3 m/4 cm, 3 m to 4 cm
Step-by-step explanation:
The actual playground will have an area of 72 m², and have one side be double the other, because the scale drawing does so too.
We can make a system of equations: l * w = 72, l = 2w
Substitute: 2w * w = 72
Simplify: 2 * w² = 72
Divide: w² = 36
Square root: <em>w = 6 m</em> (no negatives due to distances always > 0)
Substitute: l = 2 * 6 m
Simplify: <em>l = 12 m</em>
This means that the relations of one side is <em>6 m : 8 cm</em>. That can be simplified to 3 m : 4 cm.
Answer:
3rd option: B(C)= 1.79C +86.03
Step-by-step explanation:
Total bill
= cost of cans(number of cans) +cost of other groceries
Let the cost of other groceries be G, and the cost of cans be X.
Given that number of cans= C,
Total bill= XC +G
If 2 cans were purchased,
2X+G= 89.61 -----(1)
If 5 cans were purchased,
5X +G= 94.98 -----(2)
(2) -(1):
(5X +G) -(2X +G)= 94.98 -89.61
5X +G -2x -G= 5.37
3X= 5.37
X= 5.37 ÷3 <em>(</em><em>÷</em><em>3</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
X= 1.79
Subst. X= 1.79 into (1):
2(1.79) +G= 89.61
3.58 +G= 89.61
G= 89.61 -3.58 <em>(</em><em>-3.58</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em>)</em>
G= 86.03 <em>(</em><em>simplify</em><em>)</em>
Total bill
= XC +G
= 1.79C +86.03
Thus, the function is B(C)= 1.79C +86.03.
<span>8x+4y+(-8z^2)+[3z+(-5z)] the answer is D</span>
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.
Answer:
The circumference of the round pizza is 21.26in
Step-by-step explanation:
To solve this exercise we first have to calculate the radius of the circle, for this we will use the formula of area of a circle and we will clear the radius
a = area = 36in²
r = radius
π = 3.14
a = π * r²
we clear r
r = √(a/π)
now we replace the known values
r = √(36in²/3.14)
r = √(11.46in²)
r = 3.385 in
now we use the formula to calculate the circumference of a circle
c = π * 2r
c = 3.14 * 2 * 3.385 in
c = 21.26in
The circumference of the circle is 21.26in
