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3 years ago

Please solve the following quadratic equation and show your work: x^2-2x+6=0

1 answer:

6 0

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ exist\ two\ solutions\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta =0\ then\ exist\ one\ solution\ x_o=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solutions$

We have:

$x^2-2x+6=0\\\\a=1;\ b=-2;\ c=6\\\\\Delta=(-2)^2-4\cdot1\cdot6=4-24=-20 < 0$

Answer: NO SOLUTIONS.

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If x varies directly as y, then what is y when x = 4, knowing that x = 60 when y = 5?

Basile [38]

Answer: The answer would be 1/2

Step-by-step explanation:

Y varies DIRECTLY as x. Find the constant of variation, k, if y = -5x

The first sentence is there for information, the constant is a number( 1,2,3...so on) , so the only number is -5

It is a linear equation meaning equation of a straight line.

its form is y= mx+b, where m is the slope of the line, b is the y intercept when x=0. these info you have to memorize like your life depend on it.

next, y varies directly as x. If y = 15 when x = 60, find y when x = 100

if y= 15, when x=60, then there is a number m such that m*y= 60

or m= 60/y= 60/15 =4

so the equation is y=4x. in this cause, the b=0

so when x=0, y=0

The distance a car travels at a constant speed varies directly as the time the car spends traveling. If a car traveled 105 miles in 3 hours, how far could the car travel in 5 hours?

remember that distance = speed times time.

or d=Vt

so car goes 105 miles on 3 hours

how far it goes in one hour?

if you know that, then you can tell how far it goes in 5 hours.

Hope I made it non tough enough

Please give me Brainliest

4 0

3 years ago

Read 2 more answers

At an archery competition, Aylin hits the target 14 out of 20 tries in round one. If Aylin gets 24 tries in round two, approxima

Andre45 [30]

Answer:

16.8

Step-by-step explanation:

14/20 = 0.7

24 x 0.7 = 16.8

(Feel free to round up or down, or keep it the same depending on the criteria of the question!)

6 0

2 years ago

Hank is throwing a surprise party. The party costs $20 per person. Create an equation and graph to represent the relationship be

o-na [289]

Answer: Answer with Step-by-step explanation:

Since we have given that

Cost of party per person = $20.

Let the number of friends be 'f'.

Let the cost of party be 'p'.

According to question, our equation becomes

p= 2f

So, if p = 100, then f = p/2 = 100/2 = 50 if f = 10, then, p = 2f = 2 • 10 = 20 If p = 300 then, f = 300/2 = 150, If f = 20, then, p = 2f = 2 • 20 = 40 So, Table as follows:

Number of friends cost of party

50 100

10 20

150 300

20 40

3 0

2 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$