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zubka84 [21]
3 years ago
8

Multiply the polynomials (x+3)(x^2-6x+5)

Mathematics
1 answer:
LekaFEV [45]3 years ago
3 0

Answer:

\huge\boxed{x^3-3x^2-13x+15}

Step-by-step explanation:

\text{Use FOIL:}\\\\(a+b)(c+d)=ac+ad+bc+bd

(x+3)(x^2-6x+5)\\\\=(x)(x^2)+(x)(-6x)+(x)(5)+(3)(x^2)+(3)(-6x)+(3)(5)\\\\=x^3-6x^2+5x+3x^2-18x+15\\\\\text{combine like terms}\\\\=x^3+(-6x^2+3x^2)+(5x-18x)+15\\\\=x^3-3x^2-13x+15

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H(t) = -10+ - 6<br>h()= -44​
lesya [120]

Answer:

-44

Step-by-step explanation:

4 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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7) Let us use the expections formula we mentioned in 1. Substituting the possibilities and the values for all 4 events (each event is the different profit of the business at the end of the year).
E=0.2*(-10000)+0.4*0+0.3*5000+0.1*8000=-2000+0+1500+800=300$
This is the average payoff of the firm per year.
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.
8 0
3 years ago
Read 2 more answers
A poster is to have a total area of 125 cm2. There is a margin round the edges of 6 cm at the top and 4 cm at the sides and bott
jeka57 [31]

Answer:

The correct answer is 15 cm.

Step-by-step explanation:

Let the width of the required poster be a cm.

We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.

Similarly total margin combining both the sides is (4+4=) 8 cm.

So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } cm^{2}

This area is equal to 125 cm^{2} as per as the given problem.

∴ (a - 10) × (a - 8) = 125

⇒ a^{2} - 18 a +80 -125 =0

⇒ a^{2} - 18 a -45 = 0

⇒ (a-15) (a-3) = 0

By law of trichotomy the possible values of a are 15 and 3.

But a=3 is absurd as a > 4.

Thus the required answer is 15 cm.

7 0
3 years ago
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What is the speed (approximately) of a 2.5-kilogram mass after it has fallen freely from rest through a distance of 12 meters?
Norma-Jean [14]

Answer:

15.3 m/s

Step-by-step explanation:1) Find the gravitational potential energy using the equation :

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2) Then use the equation for kinetic energy to solve for the velocity:

     KE=\frac{1}{2} mv^{2}

     294=\frac{1}{2}(2.5)(v^2)\\235.2=v^2\\\sqrt{235.2} =v\\15.3 m/s=v

6 0
2 years ago
“Write an equation in slope-intercept form for the line with slope 3/2 and y-intercept -7.”
gregori [183]

Answer:

y = 3/2x - 7

Step-by-step explanation:

slope-intercept: y = mx + b

m = slope (3/2)

b = y-intercept (-7)

when put together, the equation is y = 3/2x - 7.

6 0
2 years ago
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