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3 years ago

PLEASE HELP!! Match the missing angle measurement

Mathematics

1 answer:

artcher [175]3 years ago

6 0

Answer:

FIRST TRIANGLE-

straight line under the angle always equals 180 so that is a clue to what you need to do....

180-44=136

x=136

SECOND

Since this traingle is a right angle

a right angle =90

since a line is thro it to seprate it into two angles all you need to do....

90-57=33

y=33

THIRD

it says ABC=107

witch means both angles need to add up to 107

all you do is 107-27=80

x=80

last one

This is the easiest one when a you have a intersecting angle

two numbers are the same...

so 57=p

always

So if number on top was 30

bottom angle too....

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3 years ago

Given: 4(x+3)=52 prove: x=10

Lorico [155]

Answer:

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Step-by-step explanation:

Input 10 as x,

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4 years ago

From a thin piece of cardboard 50 in. by 50 in., square corners are cut out so that the sides can be folded up to make a box. Wh

mixer [17]

Answer:

When dimension of box is 33.33 inches × 33.33 inches ×8.33 then its volume is maximum and is 9259.26 cubic inches.

Step-by-step explanation:

Let h be the length (in inches) of the square corners that has been cut out from the cardboard and that would be the height of the cardboard box.

Since the squares have been cut from cardboard, both sides of the cardboard would reduce by 2h.

Thus, The dimension of box is (50 – 2h) × (50 – 2h) × h in dimensions.

The volume V of rectangular box = (Length × Breadth × Height) cubic inches.

$V=(50-2h) \times (50-2h) \times h$

$V=(50-2h)^2 \times h$ ..............(1)

Using $(a-b)^2=a^2+b^2-2ab$

$V=h(2500+4h^2-200h)$

$V=2500h+4h^3-200h^2$

For obtaining a box of maximum volume, maximize V as a function of h.

Differentiate both sides with respect to h,

$\frac{dV}{dh}=2500+12h^2-400h$

$\frac{dV}{dh}=4(625+3h^2-100h)$

Solving quadratic equation, $625+3h^2-100h$

$\frac{dV}{dh}=4(3h^2-25h-75h+625)$

$\frac{dV}{dh}=4(h(3h-25)-25(3h-25))$

$\frac{dV}{dh}=4((h-25)(3h-25))$

For maximum, $\frac{dV}{dh}=0$

thus, $4((h-25)(3h-25))=0$

⇒ $h= 25$ or $h=\frac{25}{3}$

Now check (1) for $h= 25$ and $h=\frac{25}{3}$ .

$h= 25$ is not possible as when h is 25 inches then length and breadth becomes 0.

When $h=\frac{25}{3}$ .

(1) ⇒ $V=(50-2(\frac{25}{3}))^2 \times\frac{25}{3}=9259.2592593$

This is the maximum volume the box can assume.

Thus, when dimension of box is 33.3 inches × 33.3 inches ×8.3 then its volume is maximum and is 9259.26 cubic inches.

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3 years ago

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