All categories

3 years ago

Proportional fraction to 1/4

Mathematics

1 answer:

Pavel [41]3 years ago

7 0

A proportional fraction to 1/4

1 * 1 1
- = -
4 * 4 16

Also known as 1/16

You might be interested in

Can someone help me plz

alexandr402 [8]

Answer:

help me first plz i will help u

Step-by-step explanation:

PLEASE HELP ME

3 0

3 years ago

Read 2 more answers

Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu

kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

$f(x) = {x}^{5} - {x}^{3} + 6$

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

$f'(x) = 5 {x}^{4} - 3 {x}^{2}$

At turning point f'(x)=0.

$5 {x}^{4} - 3 {x}^{2} = 0$

This implies that,

${x}^{2} (5 {x}^{2} - 3) = 0$

${x}^{2} = 0 \: or \: 5 {x}^{2} - 3 = 0$

$x = - \frac{ \sqrt{15} }{5} \: or \: x = 0 \: \: or \: x =\frac{ \sqrt{15} }{5}$

We plug this values into the original function to obtain the y-values of the turning points

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} +750)) \:and \: (0, - 6) \: and\: ( \frac{ \sqrt{15} }{5} , \frac{1}{125} ( - 6 \sqrt{15} +750))$

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

$f''(x) = 20 {x}^{3} - 6x$

$f''( - \frac{ \sqrt{15} }{5} ) \: < \: 0$

Hence

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} + 750))$

is a maximum point.

$f''( \frac{ \sqrt{15} }{5} ) \: > \: 0$

Hence

$( \frac{ \sqrt{15} }{5} , \frac{1}{125} (- 6 \sqrt{15} + 750))$

is a minimum point.

$f''(0) \: =\: 0$

Hence (0,-6) is a point of inflexion

4 0

3 years ago

May I get some help? Thank you!

Katen [24]

Answer:

vol. of triangular prism = lbh/2

vol. of rectangular prism = 18 of vol. of triangular prism

18x = 18 + 14x

4x = 18

x=4.5 ft

so vol. of rectangular prism = 6 × 4 . 5 × 3 = 81 ftcube

7 0

3 years ago

Read 2 more answers

Finding Derivatives Implicity In Exercise, find dy/dx implicity.<br> x2 - 3 ln y + y2 = 10

Veseljchak [2.6K]

Answer:

$\frac{dy}{dx}=-\frac{2xy}{2y^2-3}$

Step-by-step explanation:

We are given that

$x^2-3lny+y^2=0$

Differentiate w.r.t x

$2x-\frac{3}{y}\frac{dy}{dx}+2y\frac{dy}{dx}=0$

By using formula

$\frac{dx^n}{dx}=nx^{n-1}$

$\frac{d(lnx)}{dx}=\frac{1}{x}$

$\frac{dy^n}{dx}=ny^{n-1}\frac{dy}{dx}$

$\frac{dy}{dx}(-\frac{3}{y}+2y)+2x=0$

$\frac{dy}{dx}(-\frac{3}{y}+2y)=-2x$

$\frac{dy}{dx}=-\frac{2x}{-\frac{3}{y}+2y}$

$\frac{dy}{dx}=-\frac{2xy}{2y^2-3}$

Hence, the derivative of function

$\frac{dy}{dx}=-\frac{2xy}{2y^2-3}$

8 0

3 years ago

A student drives 4.8-km trip to school and averages a speed of 22.6 m/s. on the return trip home, the student travels with an av

ArbitrLikvidat [17]

In this case, we cannot simply take the average speed by adding the two speeds and divide by two.

What we have to do is to calculate the time required going to school and the return trip home.

We know that to calculate time, we use the formula:

t = d / v

where,

d = distance = 4.8 km = 4800 m

v = velocity

Let us say that the variables related to the trip going to school is associated with 1, and the return trip home is 2. So,

t1 = 4800 m / (22.6 m / s)

t1 = 212.39 s

t2 = 4800 / (16.8 m / s)

t2 = 285.71 s

total time, t = t1 + t2

t = 498.1 s

Therefore the total average velocity is:

= (4800 m + 4800 m) / 498.1 s

= 19.27 m / s = 19.3 m / s

Answer:

19.3 m/s

4 0

3 years ago

Read 2 more answers