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2 years ago

?/6 < 1/2 which one is it

Mathematics

2 answers:

Alekssandra [29.7K]2 years ago

4 0

Answer:

1/6<1/2

Step-by-step explanation:

1/2 is .5 while 1/6 is .166 repeating

gayaneshka [121]2 years ago

3 0

Answer:

If 1/2 is a larger number than ?/6 the unknown number (?) has to be smaller a smaller number than 3.

Step-by-step explanation:

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Round .1325 to nearest thousandths

m_a_m_a [10]

So,

If the number in the ten-thousandth's place is greater than or equal to 5, then round up. If not, round down.

0.1325: Note the 5. We round up.

0.1325 --> 0.133

4 0

2 years ago

Read 2 more answers

Select the correct answer. What is the solution of this system of equations? y = 3x + 5 y = -x + 3

mestny [16]

Rewrite = so is on the left side.

7 0

2 years ago

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

Tresset [83]

Answer: The volume of largest rectangular box is 4.5 units.

Step-by-step explanation:

Since we have given that

Volume = $xyz$

with subject to $x+2y+3z=9$

So, let $z=\dfrac{9-x-2y}{3}$

So, Volume becomes,

$V=xyz\\\\V=xy(\dfrac{9-x-2y}{3})\\\\V=\dfrac{9xy-x^2y-2xy^2}{3}$

Partially derivative wrt x and y we get that

$9-2x-2y=0\implies 2x+2y=9\\\\and\\\\9-x-4y=0\implies x+4y=9$

By solving these two equations, we get that

$x=3,y=\dfrac{3}{2}$

So, $z=\dfrac{9-x-2y}{3}=\dfrac{9-3-3}{3}=\dfrac{3}{3}=1$

So, Volume of largest rectangular box would be

$xyz=3\times \dfrac{3}{2}\times 1=\dfrac{9}{2}=4.5$

Hence, the volume of largest rectangular box is 4.5 units.

4 0

3 years ago

1. y = x2 + 8x + 15<br> Find the zeros of the function by rewriting the function in intercept form

lubasha [3.4K]

The zeros of given function $y=x^{2}+8 x+15$ is – 5 and – 3

<u>Solution:</u>

$\text { Given, equation is } y=x^{2}+8 x+15$

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as to obtain zeros of function

We know that, intercept form of above equation is $x^{2}+8 x+15=0$

$\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3$

$\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}$

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0

2 years ago

A rectangular box is 5 in. wide, 5 in. tall, and 4 in. long. What is the diameter of the smallest circular opening through whi

Katyanochek1 [597]

Answer:

7 (7.07106781187)

Step-by-step explanation:

All you have to do is find the diameter of the rectangle from two farthest corners. To do this, use the Pythagorean Theorem

(5^2) + (5^2) = c^2

25 + 25 = c^2

50 = c^2

7 ≈ c

8 0

2 years ago