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3 years ago

The sum of the prime divisors of 2001 is a) 55, b) 56, c) 670, d) 671, e) 2001

Mathematics

1 answer:

ollegr [7]3 years ago

4 0

Answer:

Option A

Step-by-step explanation:

2001 can be divided by 3, 23, and 29 without remainders. This means that the three numbers are prime divisors of 2001.

2001/3 = 667

2001/23 = 87

2001/29 = 69

The sum of the prime divisors is 55.

3 + 23 + 29 = 55

Option A should be the correct answer.

Hope this helps.

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Alex787 [66]

Answer:

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I think this is the answer.

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2 years ago

What is the volume of a box with sidelines that are 20 inches use the formula B equals S3 where is S is the length of one side

anastassius [24]

Answer:

The volume of a box with sidelines that are 20 inches is 8000 cubic inches

Step-by-step explanation:

The volume of a box with sidelines that are 20 inches can be determined by using the formula

B = S³

Where S is the length of one side

and B is the Volume

From the question, the sidelines of the box are 20 inches. That is

S = 20 inches

From

B = S³

B = (20 inches)³

B = 20 inches × 20 inches × 20 inches

B = 8000 cubic inches

Hence, the volume of a box with sidelines that are 20 inches is 8000 cubic inches.

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2 years ago

Every Sunday John goes to church from his home . He move 2 miles east. How much distance is between home and church

GREYUIT [131]

Answer:

2 miles

Step-by-step explanation:

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8 0

3 years ago

Please help i need answer and explanation.

allsm [11]

Answer:

m<TVZ = 95

Step-by-step explanation:

2x+15+2x+5= 180

4x+20 =180

-20 -20

4x=160

x= 40

2x+15= m<TVZ

2(40)+15 = m<TVZ

80+15 = m<TVZ

95 = m<TVZ

To check:

2x+5= <WVZ

2(40)+5 = <WVZ

80+5 = <WZV

85+95 = 180

The two angles together should equal 180. Comment if you have any questions :)

6 0

2 years ago

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Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

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

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

5 0

2 years ago

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