<u>Not sure what you are asking for, but,</u>
<u>Here is an example of a JRU (Join Result Unknown) word problem</u>:
There were _____ kids on the playground. ____ more kids came onto the playground. How many kids are on the playground?
<u>Here is an example of a JCU (Join Change Unknown) word problem:</u>
There were ____ kids on the playground. Some more kids came on the playground. Now there are ____ kids on the playground. How many kids came on the playground?
<u>
Here is an example of a JSU (Join Start Unknown) word problem:</u>
Some kids were on the playground. ____ kids came on the playground. Now there are ____ kids on the playground. How many kids were on the playground at the beginning?
Answer:
72 cm
Step-by-step explanation:
Find the radius using the given information
308=1/2(πr^2)
r^2=616/π
r=14
The perimeter is πr + 2r so plug in 14
π(14) + 2(14)
43.96+28=71.96
Your answer is True!
Some other examples of an element in the set of natural numbers are 2, 3, 4, 5, etc.
Hope this helps you!
~sofia
The wire, the pole and the segment joining the leg of the pole to the wire in the ground form a right triangle whose hypotenuse is the wire, and the side opposite to the angle 36° is the pole.
By right triangle trigonometry, sin36°=(opposite side)/(hypotenuse.)
Substituting, we have 0.588=(opposite side)/220, thus the length of the opposite side, which represents the length of the pole, is
0.588*220 ft=129.3 ft
Answer:
Step-by-step explanation:
Given the formula for calculating the distance travelled expressed as;
s=1/2at^2
Given
a = 3
t = 10
Required
lower bound of s
Substitute the given values into the equation;
s=1/2at^2
S = 1/2(3)(10)^2
S = 1/2 * 3 * 100
S = 3 * 50
S = 150
Hence the lower bound of distance S is 150
