What is the solution to this equation?

Never Mind, I figured it out already

FrozenT [24]

Figured what out
a question

5 0

3 years ago

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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

7 0

2 years ago

A company makes poppy lapel pins.

mezya [45]

Answer:

<h2>The required percentage is 65.45%.</h2>

Step-by-step explanation:

The total costing is $1000 \times 1.10 = 1100$

35% were given to the charity, hence 350 were given to the charity.

There are (1000 - 350) = 650 lapel pins remaining.

60% of 650 = $\frac{60}{100} \times 650 = 390$ werre sold for 3.00 each. Hence, earning from here is $390 \times3 = 1170$ .

40% of 650 = 260 were sold at 5.00 for 2. It can also be said that $\frac{260}{2} = 130$ were sold at 5.00 each. Earning from this section is $130 \times5 = 650$

Total earning is (1170 + 650) = 1820.

Profit percentage = $\frac{1820 - 1100}{1100} \times100 = \frac{720}{11} = 65.45$

5 0

3 years ago

Say a hacker has a list of n distinct password candidates, only one of which will successfully log her into a secure system. a.

alexdok [17]

Answer:

The probability is $\frac{1}{n}$

Step-by-step explanation:

If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:

If k=1

$P = \frac{1}{n}$

Because, in her first try she has n possibles options and just one give her a successful login.

If k=2

$P=\frac{n-1}{n} *\frac{1}{n-1} =\frac{1}{n}$

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.

If k=3

$P=\frac{n-1}{n} *\frac{n-2}{n-1} *\frac{1}{n-2} = \frac{1}{n}$

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.

Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n

7 0

3 years ago

Use long division to find the quotient below (30x^3+4x^2-150) / (6x-10)

Anni [7]

Answer:

The answer is

A $.5x^2+9x+15$