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ivanzaharov [21]

3 years ago

12

Seven friends picked 7 quarts of blueberries. Three of the friends will share 4 quarts of blueberries equally and the other 4 fr

iends will share 3 quarts of blueberries equally. Which group does each friend get a greater amount of blueberries? Explain.

2 answers:

kicyunya [14]3 years ago

8 0

The 3 friends sharing 4 quarts

lesantik [10]3 years ago

6 0

The three friends sharing 4 quarts get a greater amount of blueberries because they each get over 1 quart while in the other group each friend gets less than one quart.
Hope this helps.

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3 years ago

Angle A (6×-1 B 20degrees C (×+14degrees)

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4 0

3 years ago

the diameter of a bike wheel is 25 inches. If the wheelmakes 20 Complete rotations, how far does the bike travel?

Iteru [2.4K]

Answer:

1571 in

Step-by-step explanation:

Given data

Diamter= 25 in

radius= 25/2= 12.5in

Circumference= 2πr

substitute

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C=78.55in

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4 0

3 years ago

The sum of two numbers is 95

nikklg [1K]

The sum of two number to equal 95 could equal 90 and 5

7 0

3 years ago

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the re

alukav5142 [94]

Answer:

Step-by-step explanation:

$\text{Given that:}$

$y = 1+ sec(x) \ \ y =3$

$\text{we draw the graph and the curves intersect at:}$

$x = - \dfrac{\pi}{3} \ and \ x = \dfrac{\pi}{3}$

$\text{Applying washer method;}$

$f(x) _{outer} - g(x) _{inner} --- (1)$

$V= \int ^b_a A(x) \ dx --- (2)$

$\text{outer radius = 3 - 1 = 2}$

$\text{inner radius =}$ $( 1 + sec(x) ) - 1 = sec (x)$

$A(x) = \pi ((2)^2 -(sec(x)^2) \\ \\ A(x) = \pi (4 - sec^2 (x)) ---- (3)$

$\text{The volume V =}\int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ A(x) \ dx$

$V = \int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ \pi (4- sec^2 (x) ) \ dx$

$V = 2 \pi \int ^{\dfrac{\pi}{3}}_{0}( 4 - sec^2 (x)) \ dx$

$V = 2 \pi \int ^{\pi/3}_{0} 4 . \ dx - 2 \pi \int ^{\pi/3}_{0} sec^2 (x) \ dx$

$V = 2 \pi(4) \int ^{\pi/3}_{0} 1 . \ dx - 2 \pi \Big( tan (x)\Big )^{\dfrac{\pi}{3}}_{0}$

$V = 8 \pi(x)^{\dfrac{\pi}{3}}_{0} - 2 \pi \Big( tan \dfrac{\pi}{3} -tan (0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}-{0}) - 2 \pi \Big( tan \sqrt{3}-(0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}) - 2 \pi \Big( \sqrt{3}\Big )$

$\mathbf{V = 2 \pi \Big(\dfrac{4\pi}{3}- \sqrt{3} \Big)}$

8 0

3 years ago