All categories

2 years ago

Is this statement true or false? a^(-3)+a^(-3)=a^(-6)

Mathematics

2 answers:

EleoNora [17]2 years ago

7 0

It's false

..............

prohojiy [21]2 years ago

6 0

<h3>Answer: False</h3>

Reason:

It should be a^(-3)*a^(-3)=a^(-6) based on the rule a^b*a^c = a^(b+c)

A way to see why the equation a^(-3)+a^(-3)=a^(-6) is false is to pick a number like a = 1 to find that...

a^(-3)+a^(-3)=a^(-6)

1^(-3)+1^(-3)=1^(-6)

1/(1^3) + 1/(1^3) = 1/(1^6)

1/1 + 1/1 = 1/1

1+1 = 1

2 = 1

Which is false, so this shows the original equation is false for a = 1. Therefore it is false in general.

You might be interested in

We can calculate the depth d of snow, in centimeters, that accumulates in Harper's yard during the first h hours of a snowstorm

PtichkaEL [24]

We are given an equation d = 5h, where d is the depth of snow and h is the number of hours of a snowstorm.

We need to find the number of hours of a snowstorm that accumulate the snow to 1 centimeter.

In order to find the number of hours that accumulate the snow to 1 centimeter, we need to plug d=1 in the given equation.

On plugging d=1, we get

1= 5h.

Dividing both sides by 5, we get

1/5 = h.

<h3>Therefore, it would take 1/5 hour for 1 centimeter of snow to accumulate in Harper's yard</h3>

4 0

3 years ago

Read 2 more answers

Y=3x+4<br> x+4y=-10<br> solve for substitution

maria [59]

Y=3x+4.....eq.(i) x+4y=-10....(eq.ii) putting value of y in eq.(ii),we get, x+4(3x+4)=-10 or x+12x+16=-10 or,13x=-10+16 or,13x=6 ●#x=6÷13 putting value if x in eq.(i),we get y=3(6÷13)+4 or,y=18÷13+4 or,y=(18+52)÷13 ●#y=70÷13

3 0

4 years ago

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$