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svetoff [14.1K]
3 years ago
12

The numerator of a fraction is 15 less than twice its denominator. If the numerator is increased by 5 and the denominator is inc

reased by 7, the new fraction will be equal to 2/3. What is the original fraction?
Mathematics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

  7/11

Step-by-step explanation:

Let d represent the original denominator. Then the original numerator is ...

  2d-15

The new numerator is ...

  (2d-15) +5

and the new denominator is ...

  d+7

The ratio of these is 2/3, so we have ...

  \dfrac{2d-15+5}{d+7}=\dfrac{2}{3}\\\\3(2d-10)=2(d+7) \quad\text{cross multiply}\\\\4d=44 \quad\text{add 30-2d}\\\\d=11\\\\2d-15=2(11)-15=7

The original fraction is 7/11.

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Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

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