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3 years ago

2x^3y^-3*2x^-1y^3 Someone help me with my math.

Mathematics

1 answer:

Virty [35]3 years ago

8 0

The answer is 4x^2

all you gotta do is multiply them like usually but -

(2X^3.)(2X^-1)
4X^(3-1)=4X^2

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3 years ago

SHORT RESPONSE Two angles of a triangle have measures of 35° and 80°. Find the values of the exterior angle measures of the tria

Oduvanchick [21]

Answer:

Step-by-step explanation:

Before giving the exterior angles, find the third angle. A triangle (all triangles) measure 180 degrees.

35 + 80 + x = 180

115 + x = 180

x = 180 - 115

x = 65

Exterior angles are supplements of the interior angles.

So the exterior angle to 65 degrees is 180 - 65 = 115

And the exterior angle to 35 degrees is 180-35 = 145

The exterior angle to 80 degrees is 180 - 100 = 80

5 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.