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4 years ago

If they denominator stays the same in the numerator increases what happens to the fractions ?

Mathematics

2 answers:

Montano1993 [528]4 years ago

8 0

If the numerator is bigger than the denominator the fraction become a mixed number

Example:
15/6 = 2 3/6 = 2 1/2

Oliga [24]4 years ago

4 0

The numerator increases but the denominator stays the same.

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Whats the middle number in between 12 and 17

sattari [20]

Answer:

The middle number would be (14+15)/2 = 14.5

How I found that is I did (14+15)/2.

pls mark as branliest

3 0

4 years ago

Which situation is most likely to have a constant rate of change?

lutik1710 [3]

Answer: A. Ounces of water consumed compared with the number of same-size bottles drunk

Step-by-step explanation:

When a quantity changes uniformly with respect to the change of other quantity, then we say there is a constant rate of change.

Thus, if one quantity is directly proportional to the other quantity then there is constant rate of change.

A. Ounces of water consumed is directly proportional to the number of same-size bottles drunk. Hence there is a constant rate of change of ounces water with respect to the number of same-sized bottles.

Rest of other options do not have directly proportionality relation between the quantities.

8 0

4 years ago

A pack of paper costs $3.75 including tax. Mr. Cooper wants to purchase packs of paper for his class and has a $20 budget. Write

e-lub [12.9K]

3.75p<=20

divide by 3.75

p < 5 1/3

he can purchase 5 packs of paper

7 0

3 years ago

Read 2 more answers

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3 0

3 years ago

How do I verify the identity: sec(t)+1/tan(t)=tan(t)/sec(t)-1

kompoz [17]

$\bf \textit{Pythagorean Identities}
\\ \quad \\
sin^2(\theta)+cos^2(\theta)=1
\\ \quad \\
1+cot^2(\theta)=csc^2(\theta)
\\ \quad \\
\boxed{1+tan^2(\theta)=sec^2(\theta)}\implies tan^2(\theta)=sec^2(\theta)-1\\\\
-----------------------------\\\\
\cfrac{sec(x)+1}{tan(x)}=\cfrac{tan(x)}{sec(x)-1}\\\\
-----------------------------\\\\$

$\bf \cfrac{sec(x)+1}{tan(x)}\cdot \cfrac{sec(x)-1}{sec(x)-1}\impliedby \textit{using the conjugate}\\\\
-----------------------------\\\\
recall\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
thus\qquad \cfrac{sec^2(x)-1^2}{tan(x)sec(x)-1}\implies \cfrac{sec^2(x)-1}{tan(x)sec(x)-1}
\\\\\\
\cfrac{\underline{tan^2(x)}}{\underline{tan(x)} sec(x)-1}\implies \cfrac{tan(x)}{sec(x)-1}$

4 0

4 years ago