Answer:
b = 18 m
c = 19.5 m
h = 9 m
Step-by-step explanation:
Call the smaller solid the original one, and call the larger solid the scaled one.
1. Find the scale factor
scale factor = length in scaled one / corresponding length in original one
scale factor = 7.5/5 = 1.5
The scale factor is 1.5
That means if you multiply any linear dimension of the original one by 1.5, you get the corresponding dimension of the scaled one.
Let's check our scale factor to make sure it's correct.
We see that the 5 m side corresponds to the 7.5 m side.
5 m * 1.5 = 7.5 m
Multiplying an original length by the scale factor of 1.5 gives us the scaled up length, so our scale factor of 1.5 is correct. Now we apply the scale factor of 1.5 to find b, c, and h.
b corresponds to 12 m
b = 12 m * 1.5 = 18 m
c corresponds to 13 m
c = 13 m * 1.5 = 19.5 m
h corresponds to 6 m
h = 6 m * 1.5 = 9 m
Answer:
b = 18 m
c = 19.5 m
h = 9 m
Answer: She spend $1.20 for lunch.
Step-by-step explanation:
Let the total amount be 'x'.
Half of her money spend for lunch be 
Half of her money left for a movie be
Amount she has now = $1.20
So, According to question, it becomes ,
Hence, Amount she spend for lunch is 
Therefore, she spend $1.20 for lunch.
Answer:
i believe c
Step-by-step explanation: also like the pfp
Let "a" and "b" be some number where:
a - b = 24
We want to find where a^2 + b^2 is a minimum. Instead of just logically figuring out that the answer is where a=b=12, I'll just use derivatives.
So we can first substitute for "a" where a = b+24
So we have (b+24)^2 + b^2 = b^2 +48b +576 + b^2
And that equals 2b^2 +48b +576
Then we take the derivative and set it equal to zero:
4b +48 = 0
4(b+12) = 0
b + 12 = 0
b = -12
Thus "a" must equal 12.
So:
a = 12
b = -12
And the sum of those two numbers squared is (12)^2 + (-12)^2 = 144 + 144 = 288.
The smallest sum is 288.
Answer:
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
