Mathematics, 16.04.2020 22:37 cherryy

6. ACT/SAT Tara has a cylindrical candle mold that is 8 inches high with a diameter of
3 inches. She would like to melt votive candles and reuse the wax. Each votive
candle is a sylinder with a height or 1.5 inches and a radius of 0.5 inch. How many
votive candles are needed to fill the candle mold?
Bin.
A. 12
B. 48
C. 57
D. 192
E. 226

Answers: 3

Show answers

Answers

48 (option B) votive candles are needed to fill the candle mold, If she would like to melt votive candles and reuse the wax.

Step-by-step explanation:

The given is,

Cylindrical candle mold that is 8 inches high with a diameter of 3 inches.

Votive candle is a sylinder with a height or 1.5 inches and a radius of 0.5 inch.

Step: 1

Formula to calculate the volume of cylinder,

$Volume, V = \pi r^{2} h$ .................................(1)

Where, r - Radius of cylinder

h - Height of cylinder

$\pi = 3.1415$

Step: 2

For the Candle mold,

Diameter, D = 3 inches

Radius, r = $\frac{d}{2}$

= $\frac{3}{2}$ = 1.5

r = 1.5 inches

Equation (1) becomes,

$V_{Candle mold} = \pi (1.5)^{2} (8)$

= (3.1415)(2.25)(8)

= 56.547 cubic inches

Step: 3

For the Votive candle,

r = 0.5 inches

h = 1.5 inches

Equation (1) becomes,

$V_{Candle mold} = \pi (0.5)^{2} (1.5)$

= (3.1415)(0.25)(1.5)

= 1.1780625 cubic inches

Step: 4

No. of votive candles are needed to fill the candle mold,

= $\frac{Volume_{candle mold} }{Volume _{Votive candles} }$

= $\frac{56.547}{1.1780625}$

= 48

No. of votive candles are needed to fill the candle mold = 48

Result:

48 (option B) votive candles are needed to fill the candle mold, If she would like to melt votive candles and reuse the wax.

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

Answer from: Quest

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.

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6. ACT/SAT Tara has a cylindrical candle mold that is 8 inches high with a diameter of
3 inche...

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