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EleoNora [17]
3 years ago
11

A shipping company places circular labels on all of their products. Each label measures a 4 inch radius. If the company order 10

label, how many square inches are needed? Use 3.1 for π.
Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:502.4 Square inches are need.

Step-by-step explanation:

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V square+6v-59=0<br> answer in one equation
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X=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
a=1
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c=-59
x=\frac{-6+/- \sqrt{(6)-4(1)(-59)} }{2(1)}
x=-3+ 2\sqrt{17} and -3- 2\sqrt{17}




8 0
3 years ago
Use a calculator to find the value of these data. Round the value to three
insens350 [35]

Answer:

-1, -1.1, 0.9, 1

Step-by-step explanation:

6 0
2 years ago
Norio, Gayle, and Tonya are playing a card game. They have a stack of red cards, a stack of yellow cards, and a stack of green c
Montano1993 [528]

Answer:

6

Step-by-step explanation

7 0
3 years ago
Point G is the midpoint of PQ. <br> PG equals 2X plus 3 and GQ equals 5X -5. what is X?
borishaifa [10]

Answer:

x=\frac{8}{3}=2\frac{2}{3}\approx{2.667}

Step-by-step explanation:

\overline{PG}=2x+3\\\overline{GQ}=5x-5

Since they're equidistant (equal distances), we can set them equal to one another and solve for x:

2x+3=5x-5

Adding 5 to both sides:

2x+8=5x

Subtracting 2x from both sides:

8=3x

Dividing both sides by 3:

\frac{8}{3}=x

Therefore:

x=\frac{8}{3}=2\frac{2}{3}\approx{2.667}

6 0
3 years ago
A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the
kramer

Answer:

18.73ft^3

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top=(side)^2=x^2

Area of one side face=l\times b=xy

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box=V=lbh=x^2y

Total cost=9(x^2)+5x^2+4(4xy)=14x^2+16xy

204=14x^2+16xy

204-14x^2=16xy

y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}

Substitute the values of y

V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)

Differentiate w.r.t x

V'(x)=\frac{1}{8}(102-21x^2)=0

V'(x)=0

\frac{1}{8}(102-21x^2)=0

102-21x^2=0

102=21x^2

x^2=\frac{102}{21}=4.85

x=\sqrt{4.85}=2.2

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

V''(x)=\frac{1}{8}(-42x)

Substitute the value

V''(2.2)=-\frac{42}{8}(2.2)=-11.55

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value  of x

y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft

Greatest volume of box=x^2y=(2.2)^2\times 3.87=18.73 ft^3

5 0
3 years ago
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