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tensa zangetsu [6.8K]
3 years ago
11

Determine if x-6 is a zero and find the quotiant and the remainder​

Mathematics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

○ A. No, x = -6 is not a zero of the polynomial.

The quotient is x - 29, and the remainder is 234.

Step-by-step explanation:

[x - 3][x - 20] >> Factored Form

Obviously, this is not a zero. Now, to get the remainder, we have to plug in the vertical line of <em>x = -6</em> into its conjugate, meaning an expression with opposite signs, which is <em>x + 6</em>. This is the expression we divide the dividend by, so you will have this:

\frac{{x}^{2} - 23x - 60}{x + 6}

Since the divisor is in the form of <em>x - c</em>, using Synthetic Division, we get this:

x - 29 + \frac{234}{x + 6}

You see? You have <em>x - 29</em> in the quotient, and you have 234 as the numerator remainder.

I am joyous to assist you anytime.

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Find the value of x.
dimaraw [331]

Answer:

2\sqrt{77}

Step-by-step explanation:

Use the Pythagorean theorem. a^{2} +b^{2}=c^{2}

a and b are the two side lengths

c is the hypotenuse (value across from the right angle)

Plug in the values that you are given.

a^{2} +4^{2} =18^{2}

Solve for x

x^{2} =18^{2} -4^{2}

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5 0
3 years ago
Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, on
insens350 [35]
1) We have that there are in total 6 outcomes If we name the chips by 1a, 1b, 3 ,5 the combinations are: 1a,3 \ 1b, 3 \1a, 5\ 1b, 5\ 3,5\1a,1b. Of those outcomes, only one give Miguel a profit, 1-1. THen he gets 2 dollars and in the other cases he lose 1 dollar. Thus, there is a 1/6 probability that he gets 2$ and a 5/6 probability that he loses 1$.
2) We can calculate the expected value of the game with the following: E=\frac{1}{6}*2- \frac{5}{6} *1. In general, the formula is E= \sum{p*V} where E is the expected value, p the probability of each event and V the value of each event. This gives a result of E=2/6-5/6=3/6=0.5$ Hence, Miguel loses half a dollar ever y time he plays.
3) We can adjust the value v of the winning event and since we want to have a fair game, the expecation at the end must be 0 (he must neither win or lose on average). Thus, we need to solve the equation for v:
0=\frac{1}{6}v -\frac{5}{6} =0. Multiplying by 6 both parts, we get v-5=0 or that v=5$. Hence, we must give 5$ if 1-1 happens, not 2.
4) So, we have that the probability that you get a red or purple or yellow sector is 2/7. We have that the probability for the blue sector is only 1/7 since there are 7 vectors and only one is blue. Similarly, the 2nd row of the table needs to be filled with the product of probability and expectations. Hence, for the red sector we have 2/7*(-1)=-2/7, for the yellow sector we have 2/7*1=2/7, for the purple sector it is 2/7*0=0, for the blue sector 1/7*3=3/7. The average payoff is given by adding all these, hence it is 3/7.
5) We can approach the problem just like above and set up an equation the value of one sector as an unknown. But here, we can be smarter and notice that the average outcome is equal to the average outcome of the blue sector.  Hence, we can get a fair game if we make the value of the blue sector 0. If this is the case, the sum of the other sectors is 0 too (-2/7+0+2/7) and the expected value is also zero.
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8) The firm goes even when the total profits equal the investment. Suppose we have that the firm has x years in business. Then x*300=1200 must be satisfied, since the investment is 1200$ and the payoff per year is 300$. We get that x=4. Hence, Claire will get her investment back in 4 years.
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3 years ago
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dmitriy555 [2]
This site will tell you how https://sciencing.com/write-functions-math-8315770.html
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defon

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3 years ago
Assessment items Jackie deposited $315 into a bank account that earned 1.5% simple interest each year. If no money was deposited
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The interest would be $14.17 so the total in the account is $329.17.

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