First, we look at where the equation intersects the y axis. It intersects at y = 4, which means that in the end of the equation there must be a "+4", so we can rule out the first two.

Second, we look at the slope of the line. Slope is defined as rise over run. As you can see in the graph, the line moves up 2 units while moving right 3 units. That means the coefficient of x (which is the slope) will be 2/3, which means the answer is y = 2/3x + 4.

4 0

3 years ago

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.

6 0

3 years ago