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3 years ago

Round 843,590 to the nearest thousand,

Mathematics

1 answer:

Doss [256]3 years ago

5 0

Answer:

844,000

Step-by-step explanation

Last three numbers are the hundreds, that's what you round when doing thousands. If it is higher than 500, then raise it to the next thousand, if not then keep it the same. In this case, 590>500, so I will round up to 844,000.

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6 apples and 5 oranges are shared among 3 people. If the apples are indistinguiable, oranges are also indistinguiable, how many

Mrrafil [7]

The number of ways that they can share these fruits if each person must have at least 1 fruit will be 12 ways.

<h3>How to illustrate the information?</h3>

From the information, 6 apples and 5 oranges are shared among 3 people and the apples are indistinguiable, oranges are also indistinguiable.

Therefore, the number of ways that they can share these fruits if each person must have at least 1 fruit will be:

= (6 - 3)! × (5 - 3)!

= 3! × 2!

= 3 × 2 × 2

= 12

Therefore, the number of ways that they can share these fruits if each person must have at least 1 fruit will be 12 ways.

Learn more about permutations on:

brainly.com/question/4658834

#SPJ1

3 0

1 year ago

Without solving, determine the number of real solutions for each quadratic equation.

masha68 [24]

Answer:

1. x^2 − 4x + 3 = 0

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions

Step-by-step explanation:

1. x^2 − 4x + 3 = 0

We need to compare this function with the general equation for a quadratic formula given by:

$f(x) = ax^2 + bx + c$

On this case we see that a=1, b = -4 and c =3

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

We can rewrite the expression like this:

$2n^2 +4n +2$

On this case we see that a=2, b = 4 and c =2

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

We can rewrite the expression like this:

$2x^2 +x +5$

On this case we see that a=2, b = 1 and c =5

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

We can rewrite the expression like this:

$x^2 -9x -6$

On this case we see that a=1, b = -9 and c =-6

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions

8 0

4 years ago

Factor

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(12x+13)2 hope this helps

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3 years ago

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3 years ago

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frozen [14]

Answer:

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Step-by-step explanation:

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3 years ago