All categories

3 years ago

Christian read 1/4 book every 2/3 per week. How many books does he read per week

Mathematics

2 answers:

sergiy2304 [10]3 years ago

6 0

3/8th is your answer.

He reads 3/8th books per week.

Elina [12.6K]3 years ago

3 0

Answer:

Christian reads 3/8th of a book a week.

Step-by-step explanation:

In order to find this amount, we need to multiply both numbers by a factor that will get the weeks to equal 1. In order to get 2/3 to equal 1, we need to multiply by 3/2 (the reciprocal). Now knowing this, we can multiply the number of books by that same fraction to get the amount of books in one week.

1/4 * 3/2 = 3/8

You might be interested in

5c/f=1/r, solve for r, c and f.

Nikolay [14]

Answer:

c = 1, f = 5, and r = 1

Step-by-step explanation:

5(1)/(5) = 1/(1)

5/5 = 1/1

1 = 1

8 0

3 years ago

Read 2 more answers

The ______ of the degrees of each factor is

nasty-shy [4]

Answer:

B. Sum

Step-by-step explanation:

When finding the volume, we multiplied the 3 dimensions with degree of 1, 1 and 3 and got the binominal of degree 5. We summed up the degrees: 1 + 1 + 3 = 5.

The SUM of the degrees of each factor is the degree of the product.

8 0

3 years ago

Read 2 more answers

Khi m=4 cho ma trận A= -3 -2m-2 2-2m

andreev551 [17]

Answer:

AM ONLI GR 4 SORI HEHEhehe

4 0

3 years ago

Read 2 more answers

Simplify the expression 4t + 10t

puteri [66]

Answer:

Add 4t and 10t

14t

Step-by-step explanation:

When you have the same variable you simply can add them together.

Hope this helps :)

5 0

3 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the bin is constructed from 48 square feet of sheet metal, so:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago