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3 years ago

4

Mrs. Tyson is ordering wrapping paper She can order ten 12-in by 12-in sheets for $2.95 or a roll that measures 30 in. By 2.5 ft

. For the same price. Which is a better buy?

1 answer:

prisoha [69]3 years ago

0 0

Answer:

The ten 12-in by 12-in sheets is a better buy

Step-by-step explanation:

please kindly check the attached file for explanation

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The length of a rectangle is four more than its width. If the area of the rectangle is 12, find the length of the rectangle.

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Answer:

I don’t know but I have the same question with different numbers

Step-by-step explanation:

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Read 2 more answers

The volume of a right circular cone with radius r and height h is V = pir^2h/3.

Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9 and dr = 6.8 - 5.9 = 0.9

h = 4 and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b). The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47 and dr = 6.45 - 6.47 = -0.02

h = 10 and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

8 0

3 years ago

I need help with this simplification-

astra-53 [7]

Answer:

1.31

Step-by-step explanation:

= $\frac{64}{56}$ ^2 x $\frac{1}{2.92}$ ^3 x 2.48

= 1.31 x 0.40 x 2.48

= 0.524 x 2.48

= 1.31

:)

Answer was verified and rounded

7 0

3 years ago

During the first hour of operation the farm stand owner sold 15% of the apples he had. On the second hour he sold 20% of the rem

Lelechka [254]

Answer:

200 kg

Step-by-step explanation:

First hour he sold 15% <em>(remaining is 85%)</em>

Second hour he sold 20% of remaining <em>(which is 85%)</em> -- so he sold $(0.2)(0.85)=0.17$ , which is 17% (remaining 85-17=68%)

Third hour he sold 25% of remaining <em>(which is 68%)</em> -- so he sold $(0.25)(0.68)=17$ , which is 17%

In third hour he sold 34, so 17% is 34. To find total he initially had, we need to find (100%). We can setup the ratio (letting x equal his initial amount):

$\frac{0.17}{34}=\frac{1}{x}\\0.17x=34\\x=\frac{34}{0.17}=200$

Initially the farm stand owner had 200 kg of apples.

5 0

3 years ago