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3 years ago

Please help!!!! largest, and smallest 6 digit number divisible by 2,3,4,5,6,8,9,10

Mathematics

1 answer:

Vadim26 [7]3 years ago

3 0

Answer:

Basically it has to be divisible by 8, 9, and 10 because any number that is divisible by 2, 3, 4, 5, 6 must also be divisible by 8, 9 and 10. We know that the number has to end in 0, the sum of its digits must be divisible by 9 and the last three digits must be divisible by 8. Therefore, the smallest number is 10080 and the largest is 999360.

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Do you think the equations (x−1)(x+3)=17+x and (x−1)(x+3)+500=517+x should have the same solution set? Why?

oksano4ka [1.4K]

Answer:

Those two pair of equations have the same solution set.

Step-by-step explanation:

There are two equations

(x-1)(x+3)=17+x ..... (1) and

(x-1)(x+3)+500=517+x ...... (2)

We have to check the same solution set will be there for equations (1) and (2) or not.

Now, we are going to rearrange the equation (2).

(x-1)(x+3)+500=517+x

⇒ (x-1)(x+3)=517-500+x

⇒(x-1)(x+3)=17+x

This is the same equation as equation (1).

Therefore, there will be the same solution set for equations (1) and (2). (Answer)

There are two equations

(x-1)(x+3)=17+x ..... (3) and

3(x-1)(x+3)+500=51+3x ...... (4)

We have to check the same solution set will be there for equations (3) and (4) or not.

Now, we are going to rearrange the equation (4).

3(x-1)(x+3)+500=51+3x

⇒ 3(x-1)(x+3)=3(17+x)

⇒(x-1)(x+3)=17+x

This is the same equation as equation (3).

Therefore, there will be the same solution set for equations (3) and (4). (Answer)

7 0

3 years ago

Find the surface area of the prism. 4 cm 3 cm 5 cm 9 cm

iren [92.7K]

Answer:

540cm^3

Step-by-step explanation:

4*3*5*9=12*5*9=60*9=540

8 0

4 years ago

Read 2 more answers

The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

Serhud [2]

Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

$C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi } = \sqrt{\frac{a^{2} + b^{2} }{2} }$

Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$

$\frac{C^{2} }{2(\pi )^{2} } = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} } - a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}} = b$

∴ we get

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

8 0

3 years ago

HELP HELP HELP PLSSS

FromTheMoon [43]

Answer:

option. c. 720°

.....................

5 0

3 years ago

Read 2 more answers

Choose the pair of events that is mutually exclusive.

barxatty [35]

Answer:

even numbers greater than 50 and prime numbers greater than 50

Step-by-step explanation:

An event can be considered as mutually exclusive in probability and statistics when two sample propositions cannot be held as true by any means. Therefore, the pair that clearly illustrates a mutually exclusive event would be "even numbers greater than 50 and prime numbers greater than 50."

The only prime number that is even is 2, so even numbers over 50 would be divisible by 2, and then not prime.

5 0

3 years ago