Mathematics College

Rewrite this division expression as an equivalent multiplication expression 5/8 ÷ 2/3

Answers

Answer 1

Answer:

The equivalent multiplication expression is $\frac{5}{8} \times \frac{3}{2}$

Step-by-step explanation:

Division of fractions:

A division of fractions is the same as the multiplication of the numerator by the inverse of the denominator.

5/8 ÷ 2/3

So numerator is 5/8, denominator is 2/3

$\frac{\frac{5}{8}}{\frac{2}{3}} = \frac{5}{8} \times \frac{3}{2}$

The equivalent multiplication expression is $\frac{5}{8} \times \frac{3}{2}$

Related Questions

6 Question 5 (5 points) Two masses, m and M. are placed along x-axis at a and b, respectively. Find the center of mass for the system (x, y). 9 (a+b)/2,0 12 0. (a+b)/2 15 O (am+bM)/2,0 (am+bM)/(m+M), O 18 O, (am+bM)/(m+M)

Answers

The center of mass for the system (x, y) is $\left(\frac{a m+b M}{m+M}, 0\right)$.

What is center of mass?

The unique location where the weighted relative position of the scattered mass adds to zero is known as the center of mass in physics. Here is where a force may be applied to produce a linear acceleration without also producing an angular acceleration.

The center of mass is a position defined relative to an object or system of objects. It is the average position of all the parts of the system, weighted according to their masses.

Centre of mass is

$x=\frac{m a+M b}{m+M}$

and y=0

$(x, y)=\left(\frac{a m+b M}{m+M}, 0\right)$

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Find the: x and y intercepts, asymptotes, x-coordinates of the critical points, open intervals where the function is increasing and decreasing, -coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. Using this information, sketch the graph of the function.

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Answers

The function has a relative minimum at (-1.278, -0.509) and a relative maximum at (1.278, 2.509).

How to find x-intercepts?

To find the x-intercepts, we set y = 0 and solve for x:

(x⁴/4) - x² + 1 = 0

This is a fourth-degree polynomial equation, which is difficult to solve analytically. However, we can use a graphing calculator or software to find the approximate x-intercepts, which are approximately -1.278 and 1.278.

To find the y-intercept, we set x = 0:

y = (0/4) - 0² + 1 = 1

So the y-intercept is (0, 1).

To find the vertical asymptotes, we set the denominator of any fraction in the function equal to zero. There are no denominators in this function, so there are no vertical asymptotes.

To find the horizontal asymptote, we look at the end behavior of the function as x approaches positive or negative infinity. The term x^4 grows faster than x^2, so as x approaches positive or negative infinity, the function grows without bound. Therefore, there is no horizontal asymptote.

To find the critical points, we take the derivative of the function and set it equal to zero:

y' = x³- 2x

x(x² - 2) = 0

x = 0 or x = sqrt(2) or x = -sqrt(2)

These are the critical points.

To determine the intervals where the function is increasing and decreasing, we can use a sign chart or the first derivative test. The first derivative test states that if the derivative of a function is positive on an interval, then the function is increasing on that interval. If the derivative is negative on an interval, then the function is decreasing on that interval. If the derivative is zero at a point, then that point is a critical point, and the function may have a relative maximum or minimum there.

Using the critical points, we can divide the real number line into four intervals: (-infinity, -sqrt(2)), (-sqrt(2), 0), (0, sqrt(2)), and (sqrt(2), infinity).

We can evaluate the sign of the derivative on each interval to determine whether the function is increasing or decreasing:

Interval (-infinity, -sqrt(2)):

Choose a test point in this interval, say x = -3. Substituting into y', we get y'(-3) = (-3)³ - 2(-3) = -15, which is negative. Therefore, the function is decreasing on this interval.

Interval (-sqrt(2), 0):

Choose a test point in this interval, say x = -1. Substituting into y', we get y'(-1) = (-1)³ - 2(-1) = 3, which is positive. Therefore, the function is increasing on this interval.

Interval (0, sqrt(2)):

Choose a test point in this interval, say x = 1. Substituting into y', we get y'(1) = (1)³ - 2(1) = -1, which is negative. Therefore, the function is decreasing on this interval.

Interval (sqrt(2), infinity):

Choose a test point in this interval, say x = 3. Substituting into y', we get y'(3) = (3)³ - 2(3) = 25, which is positive. Therefore, the function is increasing on this interval.

Therefore, the function is decreasing on the intervals (-infinity, -sqrt(2)) and (0, sqrt(2)), and increasing on the intervals (-sqrt(2), 0) and (sqrt(2), infinity).

To find the inflection points, we take the second derivative of the function and set it equal to zero:

y'' = 3x² - 2

3x² - 2 = 0

x² = 2/3

x = sqrt(2/3) or x = -sqrt(2/3)

These are the inflection points.

To determine the intervals where the function is concave up and concave down, we can use a sign chart or the second derivative test.

Using the inflection points, we can divide the real number line into three intervals: (-infinity, -sqrt(2/3)), (-sqrt(2/3), sqrt(2/3)), and (sqrt(2/3), infinity).

We can evaluate the sign of the second derivative on each interval to determine whether the function is concave up or concave down:

Interval (-infinity, -sqrt(2/3)):

Choose a test point in this interval, say x = -1. Substituting into y'', we get y''(-1) = 3(-1)² - 2 = 1, which is positive. Therefore, the function is concave up on this interval.

Interval (-sqrt(2/3), sqrt(2/3)):

Choose a test point in this interval, say x = 0. Substituting into y'', we get y''(0) = 3(0)² - 2 = -2, which is negative. Therefore, the function is concave down on this interval.

Interval (sqrt(2/3), infinity):

Choose a test point in this interval, say x = 1. Substituting into y'', we get y''(1) = 3(1)²- 2 = 1, which is positive. Therefore, the function is concave up on this interval.

Therefore, the function is concave up on the interval (-infinity, -sqrt(2/3)) and (sqrt(2/3), infinity), and concave down on the interval (-sqrt(2/3), sqrt(2/3)).

To find the relative extrema, we can evaluate the function at the critical points and the endpoints of the intervals:

y(-sqrt(2)) ≈ 2.828, y(0) = 1, y(sqrt(2)) ≈ 2.828, y(-1.278) ≈ -0.509, y(1.278) ≈ 2.509

Therefore, the function has a relative minimum at (-1.278, -0.509) and a relative maximum at (1.278, 2.509).

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I need this done like rn so please help!