Jose reads 45 pages of his novel in a 3 hours. At that rate, how many pages would he read in 5 hours

Mathematics

2 answers:

Len [333]3 years ago

8 0

Answer:

75 pages

Step-by-step explanation:

in 3 hours Jose reads= 45 pages

number of pages he reads in 1 hour= 45/3=15 pages

Therefore, using unitary method

number of pages he will read in 5 hours =15*5=75 pages

antiseptic1488 [7]3 years ago

5 0

Answer:

If Jose reads 45 pages in 3 hours then we need to first find how many pages he reads in 1 hour.

Step-by-step explanation:

45 page = 3 hour

Therefore, 1 hour = 45/3 = 15 pages

So, if he reads 15 pages in 1 hour he would read ,

15 x 5 = 75 pages in 5 hours

Hope I have helped you !

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Ksju [112]

Answer:

$\frac{5}{12}$

Step-by-step explanation:

The given problem is a subtraction problem between fractions. The first step in a problem like this is to convert both fractions to a common denominator. In other words, multiply both the numerator and denominator of a fraction by the same value such that the denominator of one of the fractions is the same as the other fraction. The numerator of a fraction is the value over the fraction bar, whereas the denominator is the value under it. Multiplying any fraction by a number over itself is the same as multiplying it by one, thus, the equation will remain true, and be equivalent to the original expression.

$\frac{13}{12}-\frac{2}{3}$

Convert to a common denominator,

$\frac{13}{12}-(\frac{2}{3}*\frac{4}{4})$

$\frac{13}{12}-\frac{8}{12}$

Now perform the subtraction, subtract one numerator from the other, remember, do no perfrom any more operations with the denominators,

$\frac{5}{12}$

3 0

3 years ago

The graph of g(x) is a transformation of the graph of f(x)=3^x .

marshall27 [118]

$f(x)=3^x\\\\g(x)=3^{x+a}+b\\\\\text{We have the points (-2, 2) and (-1, 4). Substitute:}\\\\\underline{+\left\{\begin{array}{ccc}2=3^{-2+a}+b\\4=3^{-1+a}+b\end{array}\right}\qquad\text{subtract both sides of the equations}\\.\ \ -2=3^{-2+a}-3^{-1+a}\\-2=3^{-2}\cdot3^a-3^{-1}\cdot3^a\\-2=\dfrac{1}{3^2}\cdot3^a-\dfrac{1}{3}\cdot3^a\\-2=\left(\dfrac{1}{9}-\dfrac{1}{3}\right)\cdot3^a\\-2=\left(\dfrac{1}{9}-\dfrac{3}{9}\right)\cdot3^a\\-2=-\dfrac{2}{9}\cdot3^a\qquad\text{multiply both sides by}\ -\dfrac{9}{2}$

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