All categories

2 years ago

Write the number for six million fifty thousand.

Mathematics

2 answers:

PtichkaEL [24]2 years ago

8 0

Answer:

6,500,000

Step-by-step explanation:

const2013 [10]2 years ago

6 0

Answer:

6,050,000

Step-by-step explanation:

You might be interested in

Solve for the proportion. x+2/5=6/15

Lelechka [254]

For this case we must solve the following proportion:

$\frac {x + 2} {5} = \frac {6} {15}$

Multiplying by 5 on both sides of the equation we have:

$x + 2 = \frac {6 * 5} {15}\\x + 2 = \frac {30} {15}\\x + 2 = 2$

Subtracting 2 from both sides of the equation:

$x = 2-2\\x = 0$

ANswer:

$x = 0$

8 0

2 years ago

Read 2 more answers

How many pairs of consecutive natural numbers have a product of less than 40000? I am in 5th grade. This is supposed to be easy

Ugo [173]

Answer:

There are 199 pairs of consecutive natural numbers whose product is less than 40000.

Step-by-step explanation:

We notice that such statement can be translated into this inequation:

$n \cdot (n+1) < 40000$

Now we solve this inequation to the highest value of $n$ that satisfy the inequation:

$n^{2}+n < 40000$

$n^{2}+n -40000$

The Quadratic Formula shows that roots are:

$n_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}$

$n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}$

$n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}$

$n_{1} \approx 199.501$

$n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}$

$n_{2} \approx -200.501$

Only the first root is valid source to determine the highest possible value of $n$ , which is $n_{max} = 199$ . Each natural number represents an element itself and each pair represents an element as a function of the lowest consecutive natural number. Hence, there are 199 pairs of consecutive natural numbers whose product is less than 40000.