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2 years ago

Write down the value of 8^-4/3

Mathematics

1 answer:

avanturin [10]2 years ago

6 0

Answer: 1/16

Step-by-step explanation:

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100 POINTS<br><br> PLEASE PROVIDE STEPS<br> THANK YOU!!

kolbaska11 [484]

Answer:

Local minimum at x = 0.

Step-by-step explanation:

Local minimums occur when g'(x) = 0 and g"(x) > 0.

Local maximums occur when g'(x) = 0 and g"(x) < 0.

Set g'(x) equal to 0 and solve:

0 = 2x (x − 1)² (x + 1)²

x = 0, 1, or -1

Evaluate g"(x) at each point:

g"(0) = 2

g"(1) = 0

g"(-1) = 0

There is a local minimum at x = 0.

5 0

3 years ago

Read 2 more answers

Ron’s class donated d dollars to a local charity. Danny’s class donated 21% more than Ron’s class. Which expression shows the am

Savatey [412]

DX1.21 because 1.21 means 21 percent more than d

6 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.