K.E=0.5mv^2
GPE=mgh
The child has both K.E and GPE
![ke = 0.5 \times 20 \times {5}^{2} \\ ke = 10 \times 25 \\ ke = 250 \: joules](https://tex.z-dn.net/?f=ke%20%3D%200.5%20%5Ctimes%2020%20%5Ctimes%20%20%7B5%7D%5E%7B2%7D%20%20%5C%5C%20ke%20%3D%2010%20%5Ctimes%2025%20%5C%5C%20ke%20%3D%20250%20%5C%3A%20joules)
![gpe = 20 \times 9.81 \times 2 \\ gpe = 40 \times 9.81 \\ gpe = 392.4 \: joules](https://tex.z-dn.net/?f=gpe%20%3D%2020%20%5Ctimes%209.81%20%5Ctimes%202%20%5C%5C%20gpe%20%3D%2040%20%5Ctimes%209.81%20%5C%5C%20gpe%20%3D%20392.4%20%5C%3A%20joules)
g is taken as 9.81
Answer:
m(arc ZWY) = 305°
Step-by-step explanation:
8). Formula for the angle formed outside the circle by the intersection of two tangents or two secants is,
Angle formed by two tangents = ![\frac{1}{2}(\text{Difference of intercepted arcs})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Ctext%7BDifference%20of%20intercepted%20arcs%7D%29)
= ![\frac{1}{2}(220-140)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28220-140%29)
= ![\frac{1}{2}(80)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2880%29)
= 40°
9). Following the same rule as above,
Angle formed between two tangents = ![\frac{1}{2}(\text{Difference of intercepted arcs})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Ctext%7BDifference%20of%20intercepted%20arcs%7D%29)
125 = ![\frac{1}{2}[m(\text{major arc})-m(\text{minor arc})]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Bm%28%5Ctext%7Bmajor%20arc%7D%29-m%28%5Ctext%7Bminor%20arc%7D%29%5D)
250 = ![[m(\text{arc ZWY})-m(\text{arc ZY})]](https://tex.z-dn.net/?f=%5Bm%28%5Ctext%7Barc%20ZWY%7D%29-m%28%5Ctext%7Barc%20ZY%7D%29%5D)
250 = m(arc ZWY) - 55
m(arc ZWY) = 305°
Therefore, measure of arc ZWY = 305° will be the answer.
10). m(arc BAC) = ![\frac{1}{2}([m(\text{arc BDC})-m(\text{arc BC})])](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Bm%28%5Ctext%7Barc%20BDC%7D%29-m%28%5Ctext%7Barc%20BC%7D%29%5D%29)
= ![\frac{1}{2}(254-106)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28254-106%29)
= ![\frac{148}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B148%7D%7B2%7D)
= 74°
Each division set gives the outcome of the operation 1.45 ÷ 5 which is 0.29.
- The number of hundredths in each division set is <u>D. 9</u>
Reasons:
The given Hunter's model consists of the following
One 10 × 10 number block
Four sets of a column of 10 cubes
Five individual cube pieces
Therefore;
In 1.45, we have;
1 unit
4 tenths
5 hundredths
Which gives;
Each single cube can be used to represent a hundredth in 0.05
One cube = 0.01
Each set of 10 cubes represents a tenth in 0.4
Each block of 10 by 10 can be used to represent the unit; 1
Dividing each of the 10 × 10 can be divided to sets of 20 blocks with a value of 0.2 each
The 4 sets of 10s can be divided by 5 to give sets of 8 with a value of 0.08
The 5 cubes divided 5 gives five cubes with each cube having a value of 0.01.
Therefore;
The value of each division set is 0.2 + 0.08 + 0.01 = 0.29
The number of hundredths in 0.29 = 9
The number of hundredths in each division set is therefore; <u>D. 9</u>
Learn more about number place value here:
brainly.com/question/184672
Answer:
2 points the first 3 matches then 3 points the last 2 matches
Step-by-step explanation:
12 points in 5 matches and the first 3 are all the same then the other 2 increased by 1 point
They got 2 points the first 3 matches then 3 points the last 2 matches
2 + 2 + 2 (first 3 matches) = 6
3 + 3 (last 2 matches) = 6
6 + 6 = 12