HURRY!

do you agree that the equation (x-4)²=9 can be solved both by factoring and extracting square roots?

zvonat [6]

Given parameters:

Equation:

(x-4)²=9

Problem, solve the equation by factoring and extracting the root

Solution:

Given equation:

(x-4)²=9

subtract 9 from both sides to equate to zero;

(x-4)² - 9 = 0

(x -4)² - 3² = 0

Now, this is similar to the concept of difference of two squares;

x² - y² = (x + y)(x-y)

Here x = x-4 and y = -3

So input and solve;

(x - 4 -3)(x - 4 -(-3)) = 0

(x - 7)(x - 4+3) = 0

(x -7)(x -1) = 0

S0,

x - 7 = 0 or x-1 = 0

x = 7 or 1

If we extract the square roots;

(x-4)² = 9

√(x-4)² = √9

x - 4 = 3

x = 4 + 3 = 7; this is not the only solution to the problem

So we cannot solve it by extracting the square root.

7 0

2 years ago

You can win concert tickets from a radio station if you are the first person to call when the song of the day is played, or if y

Anit [1.1K]

The song of the day is announced at a random time between 7:00 and 7:30, so you have a 30-minutes span of time where the song could be announced. If you start listening to the radio 20 minutes in, you're at 2/3 of the time. So, there is a 2/3 chance that the song has already been chosen, and a 1/3 chance that it will be announced in the remaining 10 minutes.

Similarly, the trivia question is asked at a random time between 7:15 and 7:45 so there is again a 30-minutes span of time. But this time, we're only 5 minutes late, so there is a 5/30=1/6 chance that the question was already asked, and a 25/30=5/6 chance that the question will be asked in the remaining 25 minutes.

So, there are four scenarios:

You have missed both the song of the day and the trivia question. Given all we said, this event has a chance of

$\dfrac{2}{3}\cdot\dfrac{1}{6} =\dfrac{2}{18}$

You miss the song, but not the trivia:

$\dfrac{2}{3}\cdot\dfrac{5}{6} =\dfrac{10}{18}$

You miss the trivia, but not the song

$\dfrac{1}{3}\cdot\dfrac{1}{6} =\dfrac{1}{18}$

You don't miss anything:

$\dfrac{1}{3}\cdot\dfrac{5}{6} =\dfrac{5}{18}$

You're interested in the probability of missing either of the announcements, so you want to sum the two middle cases (in the first you miss everything, in the second you don't miss anything).

Your result is thus

$\dfrac{10}{18}+\dfrac{1}{18} = \dfrac{11}{18}$

3 0

2 years ago

A local animal feed company makes its feed by the ton, which is 2000 pounds. They want to include a medication in the feed. Each

pshichka [43]

100 milligrams for every 5 pounds

2000/5=400 doses

400*100=40,000 milligrams= 40 grams

7 0

2 years ago

What are the coefficients in this expression? 6a-3/4+7.5-b/3

solniwko [45]

To solve this problem you must apply the proccedure shown bellow:

1- You have the following expression given in the problem above:

$6a-\frac{3}{4}+7.5-\frac{b}{3}$

2- If $a$ and $b$ are variables, simplify the expression by adding the constants, which are the numbers without a letter:

$6a-\frac{b}{3}+6.75$

3. By definition, the coefficients are the numbers that are multiplying a variable. Therefore the coefficients are: $6$ and $\frac{1}{3}$

3 0

2 years ago

A car dealer recommends that transmissions be serviced at 30,000 miles. To see whether her customers are adhering to this recomm

salantis [7]

Answer:

No, the owners are not having their transmissions serviced at 30,000 miles.

Step-by-step explanation:

We are given that a car dealer recommends that transmissions be serviced at 30,000 miles.

The car dealer selects a random sample of 40 customers and finds that the average mileage of the automobiles serviced is 30,456. The standard deviation of the population is 1684 miles.

Let $\mu$ = true average mileage of the automobiles serviced.

So, Null Hypothesis, $H_0$ : $\mu$ = 30,000 miles {means that the owners are having their transmissions serviced at 30,000 miles}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 30,000 miles {means that the owners are having their transmissions serviced at different than 30,000 miles}

The test statistics that will be used here is One-sample z-test statistics because we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average mileage serviced = 30,456 miles

$\sigma$ = population standard deviation = 1684 miles

n = sample of customers = 40

So, the test statistics = $\frac{30,456-30,000}{\frac{1684}{\sqrt{40} } }$

= 1.71

The value of z-statistics is 1.71.

Also, the P-value of the test statistics is given by;

P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.9564 = 0.0436

For the two-tailed test, the P-value is calculated as = 2 $\times$ 0.0436 = 0.0872.

Since the P-value of our test statistics is less than the level of significance as 0.0872 < 0.10, so we have sufficient evidence to reject our null hypothesis as the test statistics will fall in the rejection region.

Therefore, we conclude that the owners are having their transmissions serviced at different than 30,000 miles.

4 0

2 years ago