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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.

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

Tema [17]

X= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
a=1
b=6
c=-59
x= $\frac{-6+/- \sqrt{(6)-4(1)(-59)} }{2(1)}$
x= $-3+ 2\sqrt{17}$ and $-3- 2\sqrt{17}$

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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:

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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

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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