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3 years ago

Help Please...

Mathematics

1 answer:

ycow [4]3 years ago

6 0

11 5/8 in = 11.625 in

9 3/4in = 9.75 in

10 1/2 in= 10.5 in

11.625+9.75+10.5= 31.875in

To find the average divide the total of inches by the amount of pieces there are to find the average:

31.875 / 3 = 10.625in

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(3cosx-4sinx) + (3sinx+4cosx) = 5

Ksenya-84 [330]

(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0

Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º (360º-53.13º=306.87)

sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º

The solutions when 0≤x≤360º are: 36.87º and 306.87º.

3 0

2 years ago

Please help me with this one.

borishaifa [10]

Answer:

V = 16
W = 48

Step-by-step explanation:

The triangles are congruent when corresponding sides are congruent.

ΔFGH ≅ ΔABC means corresponding sides are ...

(GH, BC) = (V, 16)

(HF, CA) = (48, W)

Corresponding sides being congruent means ...

V = 16

W = 48

5 0

2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

3 years ago

Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 132,127,106,140,158,135,129,138 med

Ket [755]

Answer:

133.5, 127.5, 139.5, 12

Step-by-step explanation:

order data:

106, 127, 129, 132, 135, 138, 140, 158

Median:

The middle number: (8+1)/2 = 4.5 between the 4&5 numbers

= (135-132)/2= 1.5

1.5 + 132 = 133.5

lower quartile (1st quartile):

(8+1)/4 = 2.25 between the 2&3 numbers

(129+127)/4=0.5

0.5+127 = 127.5

upper quartile(3rd quartile):

(8+1)/4 x3 = 6.75 between the 6&7 numbers

(140-138)/4 x3 = 1.5

1.5 + 138 =139.5

Interquartile range:

139.5-127.5= 12

hope this helps

4 0

3 years ago

Suppose your school plans a musical. The directors goal is ticket sales of at least $4500. Adult tickets are $7.50 and student t

Evgesh-ka [11]

D because it says you will be getting at least 4500 or more dollars by adding the adult and student tickets

7 0

3 years ago