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3 years ago

Write a fraction that is less then 5/6 and has a denominator of 8

2 answers:

8 0

So we want to write a fraction that is less than 5/6 and has a denominator of 8. So lets start from 1/1 that is greater then 5/6 and doesn't have 8 as a denominator. Lets take one half from 1/1 and well get 1/2.

rewona [7]3 years ago

3 0

1/8?

hope this helps

can you help me with a question

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True [87]

The answer is c) F and H

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3 years ago

Read 2 more answers

What is the square root of 5 divided by the square root of 15 and simplify and in fraction form

ozzi

Answer:

$\sqrt{\frac{1}{3} }$

$\frac{\sqrt{3} }{3}$

Step-by-step explanation:

The expression $\frac{\sqrt{5} }{\sqrt{15} }$ can be simplified by first writing the fraction under one single radical instead of two.

$\frac{\sqrt{5} }{\sqrt{15} } = \sqrt{\frac{5}{15} }$

5/15 simplifies because both share the same factor 5.

It becomes $\sqrt{\frac{5}{15} } = \sqrt{\frac{1}{3} }$

This can simplify further by breaking apart the radical.

$\sqrt{\frac{1}{3} } = \frac{\sqrt{1} }{\sqrt{3} } = \frac{1}{\sqrt{3} }$

A radical cannot be left in the denominator, so rationalize it by multiplying by √3 to numerator and denominator.

$\frac{1}{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } =\frac{\sqrt{3} }{3}$

4 0

3 years ago

Given the following inverse variation find the missing value: (5,-8) (20, y) find the missing value

forsale [732]

This answer is pretty simple. you see (5,-8) and (20,y). well the 20 is 4 times the 5 on the x value so multiply the y value by 4 to get your answer which Y=-32 in the 2nd corrdenants.

6 0

3 years ago

How to solve the function 4y+x=12

Sidana [21]

First you rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation

Second you solve it by using the formula of a straight line drawn on Cartesian coordinate system in which “y” is the vet risk axis and “x” the horizontal axis.

7 0

2 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

3 years ago