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Triss [41]
3 years ago
5

Which calculation will ALWAYS give a result greater than 1? A.5 × a number less than 1 B.4/9+ a fraction less than 12 C. 1 3/4 -

a fraction less than 3/4 D.7/8 × a number less than 1
Mathematics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

  C.  1 3/4 - (a fraction less than 3/4)

Step-by-step explanation:

Your number sense should be able to help you with this one.

A. 5 × 1/10 = 1/2, not a number greater than 1

B. 4/9 + 1/3 = 7/9, not a number greater than 1

C. 1 3/4 -1/2 = 1 1/4, a number greater than 1 (see below for more explanation)

D. 7/8 × 1/2 = 7/16, not a number greater than 1.

__

<em>More explanation</em>

Let x represent a number less than 3/4. Then we want to make sure that ...

  y = 1 3/4 - x

will be greater than 1.

Solving for x, we get ...

  x = 1 3/4 - y

Applying the requirement that x < 3/4, we have ...

  x < 3/4

  (1 3/4 -y) < 3/4

  1 3/4 < y + 3/4 . . . . . . add y

  1 < y . . . . . . . . . . . . . . .subtract 3/4

We see that the condition on x makes sure that y is always greater than 1.

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Can show me how to get this answer?
Olegator [25]
What you are given is f(x)=4x-8. When it says f(16)-f(0), it is telling you what the values of x should be equal to when placed into the given function. Whenever you see an x in your given function, it should be replaced with the number in parentheses after x. In this case, what you want to know is x=16 and x=0 since these are the numbers following f.
To get to your answer of 64, start out by setting up your equations.
Your first would be 4(16)-8 (since you’re filling 16 in for x).
Your next would be 4(0)-8 (since you’re filling 0 in for x).
Then you want to subtract these 2 equations from one another, as that’s what it’s asking for in the problem AKA f(16)-f(0).
4•16=64. 64-8= 56.
4•0=0. 0-8= -8.
56-(-8)= 64 since 2 negatives equal a positive.
Hope this helped! Please message me if you want to see the problem done out on a piece of paper or if you have any more questions!
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