Find all the values of k so that the given points are 29 square units apart. (-5, 5) and (k,0).

1 answer:

7 0

Distance formula between 2 points (x1,y1) and (x2,y2)

D= $\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}$

D=29 and one point is (-5,5) and the other is (k,0)
sub

29= $\sqrt{(-5-k)^{2}+(5-0)^{2}}$
29= $\sqrt{(-5-k)^{2}+(5)^{2}}$
square both sides
841= $(-5-k)^{2}+(5)^{2}$
841= $(-5-k)^{2}+25$
minus 25 both sides
816= $(-5-k)^{2}$
square root both sides don't forget positive and negative roots
+/-4√51=-5-k
add 5 to both sides
5+/-4√51=-k
times -1 both sides
-5+/-4√51=k

the possible values for k are
k=-5+4√51 or k=-5-4√51

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

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Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$