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3 years ago

Rewrite the equation by completing the square. x^2+ 12x + 36 = 0

Mathematics

2 answers:

Maslowich3 years ago

5 0

Answer:

(x+6)^2 =0

Step-by-step explanation:

svetoff [14.1K]3 years ago

4 0

Answer:

(x+6)^2 =0

Step-by-step explanation:

x^2+ 12x + 36 = 0

Subtract 36 from each side

x^2+ 12x + 36-36 = 0-36

x^2 +12x = -36

Take the coefficient of x

Divide by 2

12/2 =6

Square it

6^2 =36

Add this to each side

x^2+ 12x + 36 = -36+36

(x+12/2)^2 = 0

(x+6)^2 =0

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Financial advisors counsel young people to start saving early for their retirement. Explain why an early start is important.

Inessa05 [86]

Answer:

The more money you invest and the earlier you start means your retirement savings will have that much more time and potential to grow and investing early you can be able to take advantage of compound earnings.

Step-by-step explanation:

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3 years ago

Jason and his children went into a grocery store and will buy bananas and mangos. Each banana costs $0.90 and each mango costs $

Andrej [43]

Answer:

25 ≥ 0.90b + 2m

Step-by-step explanation:

Given:

Cost of each banana = $0.90

Cost of each mango = $2

Total amount = $25

Find:

Inequality

Computation:

Assume;

Number of banana = b

Number of mango = m

Total amount ≥ Cost of total banana + Cost of total mango

25 ≥ 0.90b + 2m

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3 years ago

If a population of 300 cells doubles every hour, which function represents the population after (h) hours

tangare [24]

I say y=300(2x) X=hours Y=population
Y=600x

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3 years ago

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0

2 years ago

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce

LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 kg.

92.82% of the sample means are between 75 kg and 83 kg.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be 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.

In this problem, we have that:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that $\mu = 79, \sigma = 22$ .

What percentage of individual adult females have weights between 75 kg and 83 kg?

This percentage is the pvalue of Z when $X = 83$ subtracted by the pvalue of Z when $X = 75$ . So:

X = 83

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

$Z = \frac{83 - 79}{22}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714.

X = 75

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

$Z = \frac{75- 79}{22}$

$Z = -0.18$

$Z = -0.18$ has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?

Now we use the Central Limit THeorem, when $n = 100$ . So $s = \frac{22}{\sqrt{100}} = 2.2.$