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3 years ago

Wilson paints 40% of a bookcase in 20 minutes how much more time will it take him to finish the bookcase?

Mathematics

2 answers:

Nadya [2.5K]3 years ago

5 0

Answer:

30 minutes

Step-by-step explanation:

First you can break down the percent and the time into a fraction

You can divide it by 2

so 2%/1 (2% in 1 minute)

Then you find the amount of time it would take to paint the whole bookcase

2 * 100 = 50

So time to paint the whole bookcase would be 50

50 - 20 = 30

larisa [96]3 years ago

4 0

Answer:

30 minutes

Step-by-step explanation:

He paints 40% of the bookcase in 20 minutes.

The rest of the bookcase is 100% - 40% = 60%.

60% of the bookcase still needs painting.

40% is to 20 minutes as 60% is to x minutes

40/20 = 60/x

Reduce the fraction on the left side.

2 = 60/x

Multiply both sides by x.

2x = 60

Divide both sides by 2.

x = 30

Answer: 30 minutes

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(2/3) (strawberries / minutes) * (60/1) (minutes / hours) = (40 strawberries / hours)
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Answer:

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Step-by-step explanation:

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The scale that will produce the smallest drawing would be b) 1mm:50m.

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b) 0.01cm:5000cm
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d) 10cm:2500cm

Then, you divide the scales:

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b) 0.000002
c) 0.005
d) 0.004

B is the smallest number, meaning it will be the smallest scale.

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3 years ago

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Answer:

Step-by-step explanation:

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3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers