Answer:
D. b = r+z
Step-by-step explanation:
Given the expression b-r = z, we are to solve for b. To do this, we will add 'r' to both sides of the equation as shown;
b-r+r = z+r
Since -r+r = 0, substitute:
b+0 = z+r
b = z+r
Hence the resulting equation when r is added to both sides of the equation is b = z+r
Hence option D is correct
Answer:
162 metres
Step-by-step explanation:
Since h is proportional to the square of v, we know that their ratio must be constant, so 
where v1 and v2 are velocities and h1 and h2 are their respective heights.
Since we are given that v = 10 and h = 8, we can set v1 = 10 and h1 = 8 and since we are trying to find the height for v = 45, we can set v2 = 45. Inputting these values into the equation and solving, we get
10^2/8 = 45^2/h2
h2 = 45^2/(10^2/8) = 162 metres
I hope this helps!
Answer:
(x²+9x)/4x
Step-by-step explanation:
hello :
(x+9/x) / (4/x) = (x+9/x)×(x/4) = x(x+9)/4x
(x+9/x) / (4/x) =(x²+9x)/4x
Answer:
4 4 0 1
Step-by-step explanation:
Q1. 4
Q2. 4
Q3. 0
Q4. 1
Answer:
1a) 210
1b) 0.18
2) 17.7 will fit (> 17)
3) 515 cm^2
Step-by-step explanation:
1a) The sample of 25 is exactly 1/30 of the population of 750, so the number of orange cakes in the population will be 30 times the number in the sample.
The total number of orange cakes made on Monday is 30·7 = 210.
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1b) The lowest number that will round to 5 is 4.5. The lowest probability that a cake is an orange cake is ...
4.5/25 = 18/100 = 0.18
The lower bound on the probability that the cake is orange is 0.18.
____
2) The working shown is partly correct.
The possible error in the box dimension is 1/2 cm, so the box may be as short as 28 -1/2 cm = 27.5 cm.
The possible error in the disc case dimension is 1/2 mm = 0.05 cm, so the case may be as thick as 1.5 +.05 = 1.55 cm.
The minimum number discs that will fit in the case is (27.5 cm)/(1.55 cm) = 17.7. Thus, 17 cases will definitely fit in the box.
____
3) The shaded area will be the largest when the outside rectangle is the largest and the inside rectangle is the smallest. Errors in dimensions can be as much as 1/2 cm, so the Outside rectangle can be 42.5 cm by 21.5 cm. The inside rectangle can be as small as 27.5 cm by 14.5 cm.
Then the shaded area can be as large as ...
(42.5)(21.5) -(27.5)(14.5) = 515 . . . . cm^2
