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mariarad [96]
3 years ago
12

3 1/4 cups - 1/4 of a cup=

Mathematics
2 answers:
elena55 [62]3 years ago
8 0

Answer:

3 cups

Step-by-step explanation:

3 1/4 - 1/4 = 3

So, your answer would be 3 cups.

nika2105 [10]3 years ago
3 0

Answer:

3 cups

Step-by-step explanation:

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Katarina [22]
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In order to accumulate enough money for a down payment on a​ house, a couple deposits $608 per month into an account paying 6% c
____ [38]
Hi there
The formula of the future value of annuity ordinary is
Fv=pmt [(1+r/k)^(kn)-1)÷(r/k)]
Fv future value?
PMT monthly payment 608
R interest rate 0.06
K compounded monthly 12
N time 6years
So
Fv=608×(((1+0.06÷12)^(12×6)
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8 0
3 years ago
a box of crackers has a volume of 48 cubic inches. what is the volume of a similar box that is smaller by a scale factor of 2/3
earnstyle [38]
For this case what you need to know is that the original volume of the cookie box is:
 V = (w) * (l) * (h)
 Where,
 w: width
 l: long
 h: height.
 We have then:
 V = (w) * (l) * (h) = 48 in ^ 3
 The volume of a similar box is:
 V = (w * (2/3)) * (l * (2/3)) * (h * (2/3))
 We rewrite:
 V = ((w) * (l) * (h)) * ((2/3) * (2/3) * (2/3))
 V = (w) * (l) * (h) * ((2/3) ^ 3)
 V = 48 * ((2/3) ^ 3)
 V = 14.22222222 in ^ 3
 Answer:
 the volume of a similar box that is smaller by a scale factor of 2/3 is:
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6 0
3 years ago
Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
Darya [45]

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

3 0
3 years ago
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