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3 years ago

(1) Two segments intersect in one point. (2) The segments are in the same plane. (3) The segments are perpendicular.

Mathematics

1 answer:

andreev551 [17]3 years ago

8 0

These should be called perpendicular bisectors

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Divide the following fractions 1/4÷1/2

artcher [175]

1/4 / 1/2 =
1/4 x 2/1 = 2/4 = 1/2

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3 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

Find f(4) for the function below. <br><br> f(x) = |x-7| + 3

Anastaziya [24]

Step-by-step explanation:

f(x) = |x-7| + 3

when x is equal to 4, then

f(4) = |4-7| + 3

=-3+3

6 0

3 years ago

Read 2 more answers

One side of a rectangle is 9 millimeters shorter than six times the other side. Find the length of the longer side if we also kn

vodka [1.7K]

Given:

W(width) = (6L) - 9

L(length) = L

Equation:

2( [ 6L ] - 9) + 2 (L) = 150

= 12L - 18 + 2L = 150

= 12L + 2L = 150 + 18

=14L = 168

L = 168/14, so the length is 12. Let's check our work.

Width: 6(12) - 9 = 72 - 9 = 63

Length: 12

Since there are two lines of width and two lines of length:

2(12) + 2(63) = 24 + 126, which gives you a perimeter of 150 mm.

Hope this helped.

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3 years ago

Find the exact value of the following.

qwelly [4]

The answer is going to be 11.98. hope that helped

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3 years ago