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3 years ago

Plz help math #5 questions

Mathematics

1 answer:

vekshin13 years ago

5 0

Okay , wheres the questions at tho...?

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A milligram is what fraction of a kilogram

pishuonlain [190]

The answer is 1/1000000

4 0

2 years ago

Judy ran the distance s. Monday 0.35, Wednesday 0.2 and Friday 0.25. write the total distance in miles as a fraction in simplest

Alexandra [31]

Answer:

4/5

Step-by-step explanation:

0.35 + 0.2 + 0.25 = 0.80

3 0

2 years ago

The denominator of a fraction is 4 more than the numerator if both are decreased by 3 the simplified result is 6/7 find original

Hitman42 [59]

Answer:

The denominator of a fraction is 4 more than the numerator. The original fraction is $\frac{27}{31}$

Solution:

Given that

Denominator of fraction is 4 more than the numerator.

Let’s say numerator of fraction be represented by variable x.

So denominator of a faction as it is four more that numerator will be x + 4

Also given if both decreased by three than simplified result is $\frac{6}{7}$

$=>\frac{(x-3)}{((x+4)-3)}=\frac{6}{7}$

Solving above equation for x

=> 7(x – 3 ) = 6 ( x + 1 )

=> 7x – 21 = 6x + 6

=> 7x – 6x = 6 + 21

=> x = 27

Numerator of fraction = x = 27

Denominator of fraction = x + 4 = 27 + 4 = 31

$\text {required fraction}=\frac{\text {numerator}}{\text {denominator}}$

$= \frac{27}{31}$

Hence the original fraction is $\frac{27}{31}$

7 0

3 years ago

A rectangle has an area of 40 square units. The length is 6 units greater than the width.

stiks02 [169]

Answer:

Answer:

The width is 4 units, and the length is 10 units.

Step-by-step.

Step-by-step explanation:

area of rectangle = length * width

Let L = length; let W = width.

"The length is 6 units greater than the width.": L = W + 6

area = LW = 40

Since L = W + 6, we substitute L with W + 6.

(W + 6)W = 40

W^2 + 6W = 40

W^2 + 6W - 40 = 0

(W - 4)(W + 10) = 0

W - 4 = 0 or W + 10 = 0

W = 4 or W = -10

A width cannot be a negative number, so we discard the solution W = -10.

W = 4

L = W + 6 = 4 + 6 = 10

The width is 4 units, and the length is 10 units.

3 0

2 years ago

Read 2 more answers

Find the global maximum and global minimum values (if they exist) of x 2 + y 2 in the region x + y = 1. If there is no global ma

Grace [21]

Given that $x+y=1$ , we have $y=1-x$ , so that

$f(x,y)=x^2+y^2\iff g(x)=x^2+(1-x)^2$

Take the derivative and find the critical points of $g$ :

$g'(x)=2x-2(1-x)=4x-2=0\implies x=\dfrac12$

Take the second derivative and evaluate it at the critical point:

$g''(x)=4>0$

Since $g''$ is positive for all $x$ , the critical point is a minimum.

At the critical point, we get the minimum value $g\left(\frac12\right)=f\left(\frac12,\frac12\right)=\frac12$ .

5 0

2 years ago