What are the fraction of the shape shaded on these three shapes?

there are 90 people in a group. 6 of the people are men. The number of women is 2 times the number of men. the rest are children

stepladder [879]

Answer:

n= 12

Step-by-step explanation:

6*2= 12

Men= 6

Women= 12

12+6= 18

90-18= 72

72:6= 12:1

3 0

2 years ago

GEOMETRY HELP ASAP I'M FAILING. I did the first half now i cant figure this out!!?

Shalnov [3]

OK, so;
BDE and BED are congruent because the opposite sides are both congruent
To find BDE and BED you must subtract 66 degrees from 180 degrees.
You are then left with 114 as the sum of both the angles you need to find
Since they are congruent, all you need to do is divide by two
114/2=57 degrees for both BDE (a) and BED(b)
Now for angle A and C;
This is easy because they are both congruent to the first two!
So basically, all of question four is "57 degrees"
Sadly for number 5 i did not understand the question :"(
For 6 tho;
AC is parallel to DE because angle C is congruent to angle BED
All the others can be ruled out
For 7;
BD is half the length of AE, so:
4x+20=2(3x+5)
4x+20=6x+10
20=2x+10
10=2x
x=5
This means BD is 20 bc
3(5)+5
15+5
20
And AE is 40 bc
20X2=40
or...
4(5)+20

5 0

2 years ago

A shoe store displayed 75 pairs of shoes on 12 shelves. How many pairs of shoes could the store display on eight shelves.

yanalaym [24]

50 pairs of shoes since 75/12 is 6.25 and then multiply 6.25x8

8 0

3 years ago

Read 2 more answers

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$