Answer— 45/15=3 so there is 3 classes with 15 students in each class? I didn’t really understand your question but I tried!
Hope it helped!
2√20=2((√4)(√5)=2((2)(√5)=4√5
now we have
4√5-3√5=1√5=√5
From the Venn diagram: 15 players like Chemstrand, 17 players like Chemgrass, 13 players like both Chemstrand and Chemgrass while 10 players like neither Chemstrand nor Chemgrass.
The missing values in the frequency table are x - representing the number of players that like both Chemstrand and Chemgrass, y - representing the number of players that like Chemgrass but do not like Chemstrand and z - representing the number of players likes Chemstrand but do not like Chemgrass.
The number of players that like both Chemstrand and Chemgrass is 13. The number of players that like Chemgrass but do not like Chemstrand is 17. The number of players likes Chemstrand but do not like Chemgrass is 15.
Therefore, x = 13, y = 17 and z = 15
<h2>Question 9:</h2>
1. Use Pythagorean Theorem (a²+b²=c²) to solve for missing side of triangle and rectangle. x²+16²=20², or x²+256=400. So, x²=144, and x=12
2. Use formula: 1/2(h)(b1+b2). 1/2 (12) (30+14).
3. Simplify: 1/2 (12) (44)=1/2(528)=264
Area of whole figure is 264 square mm.
<h2>Question 10:</h2>
Literally same thing but with trigonometry.
1. Use sine to find out length of dotted line: sin(60°)=x/12
2: Simplify: 12*sin(60°)=x. x≈10.4 (rounded to the nearest tenth)
3. Use Pythagorean Theorem to find out last leg of triangle: 10.4²+x²=12²
4: Simplify: 108.16 +x²=144. x²=35.84 ≈ 6
5: Use formula: 1/2(h)(b1+b2). 1/2 (10.4) (30+36)
6: Simplify: 1/2 (10.4) (66) =343.2
7: Area of figure is about 343.2
Remember, this is an approximate answer with rounding, so it might not be what your teacher wants. The best thing to do is do it yourself again.
