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2 years ago

If a = (3-2√2) find the value of a4+ 1/a4

Mathematics

1 answer:

FinnZ [79.3K]2 years ago

4 0

Answer:

1154

Step-by-step explanation:

Perhaps the easiest way to evaluate this numerical expression is to let a calculator do it. Alternatively, we can compute the value from a +1/a.

<h3>expression</h3>

Consider the square of a +1/a:

(a +1/a)^2 = a^2 +2(a)(1/a) +1/a^2 = (a^2 +1/a^2) +2

This means ...

a^2 +1/a^2 = (a +1/a)^2 -2

Similarly, using a^2 for 'a' in the above, we have ...

a^4 +1/a^4 = (a^2 +1/a^2)^2 -2

<h3>numerical value</h3>

The value of a +1/a is ...

$a+\dfrac{1}{a}=3-2\sqrt{2}+\dfrac{1}{3-2\sqrt{2}}=(3-2\sqrt{2})+\dfrac{3+2\sqrt{2}}{3^2-(2\sqrt{2})^2}\\\\=(3-2\sqrt{2})+(3+2\sqrt{2})\\\\a+\dfrac{1}{a}=6$

Then the value of a^2 +1/a^2 is 6^2 -2 = 34

and the value of a^4 +1/a^4 is 34^2 -2 = 1154

You might be interested in

Express 6^1/4 b^3/4 c^1/4 using a radical

hodyreva [135]

Answer:

Step-by-step explanation:

$6^{\frac{1}{4} } b^{\frac{3}{4} }c^{\frac{1}{4} }\\\\=(6^1b^3c^1)^{\frac{1}{4} }\\\\=(6b^3c)^\frac{1}{4} \\\\=\sqrt[4]{6b^3c}$

so answer is B

8 0

3 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

3 0

3 years ago

Read 2 more answers

Suppose the heights of a population of people are normally distributed with a mean of 65.5 inches and a standard deviation of 2.

kupik [55]

Answer:

a) $P(64.2$

b) 69.764

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(65.5,2.6)$

Where $\mu=65.5$ and $\sigma=2.6$

We are interested on this probability

$P(64.2$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(64.2$

And we can find this probability on this way:

$P(-0.50$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.50$

3) Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.05$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

If we use condition (b) from previous we have this:

$P(X$