Answer:
A
Step-by-step explanation:
i took the edgenuity.com Real-World Decimal Problems with Multiplication and Division Quiz it said it was right, good luck :P
Answer:
80 or 80.3
Step-by-step explanation:
You have to find what is 55% of 146
first set up your fractions. 
=
for the fraction on the left the denominator is the whole amount (146) and n is the unknown number
For the fraction on the right, the numerator is where the percent goes (55) and the denominator is always 100
Next you multiply using the butterfly method
you multiply across. 146 x 55, which is 8030, and 100 x n, which is 100n,
Now divide 8030 by 100
All you have to do is move the decimal two places, since there are two zero's, and you get 80.3
When you round 80.3 you get 80
hope this helps :)
M(x) = 4x^3 - 5x^2 - 7x
Let us first find the zeros of the function.
That is when it is equal to zero.
m(x) = 4x^3 - 5x^2 - 7x = 0
x(4x^2 - 5x - 7) = 0. Therefore x = 0 or 4x^2 - 5x - 7 = 0.
Using a quadratic function calculator to solve 4x^2 - 5x - 7
x = 2.09, -0.84
Therefore the zeros are x =-0.84, 0, 2.09 for the function m(x).
The intervals observed are imagining that the zeros are on the number line:
x<-0.84, -0.84<x<0, 0<x<2.09, x>2.09.
For each of this range we would test the function with a number that falls in the range.
The function is decreasing in the interval where it is less than 0.
For x<-0.84, let us test x = -1, m(x) = 4x^3 - 5x^2 - 7x = 4(-1)^3 - 5(-1)^2 - 7(-1) = -4 -5 +7 = -2, -2 < 0, so it is decreasing here.
For -0.84<x<0, let us test x = -0.5, m(x) = 4x^3 - 5x^2 - 7x = 4(-0.5)^3 - 5(-0.5)^2 - 7(-0.5) = -0.5 -1.25 +3.5 = 1.75, 1.75 >0. It is not decreasing.
For 0<x<2.09, let us test x = 1, m(x) = 4x^3 - 5x^2 - 7x =
4(1)^3 - 5(1)^2 - 7(1) = 4 -5 -7 = -8, -8 <0. It is decreasing.
For x>2.09, let us test x = 3, m(x) = 4x^3 - 5x^2 - 7x =
4(3)^3 - 5(3)^2 - 7(3) = 108 -45 -21 = 42, 42 >0. It is not decreasing.
So the function is decreasing in the intervals:
x < -0.84, & 0<x<2.09.
Answer: Conjecture: There is no triangle with side lengths N, 2N, and 3N (where N is a positive real number)
Proof:
We prove this by contradiction: Suppose there was an N for which we can construct a triangle with side lengths N, 2N, and 3N. We then apply the triangle inequalities tests. It must hold that:
N + 2N > 3N
3N > 3N
3 > 3
which is False, for any value of N. This means that the original choice of N is not possible. Since the inequality is False for any value of N, there cannot be any triangle with the given side lengths, thus proving our conjecture.
