All categories

3 years ago

B. Why does using FOIL on polynomial expressions match so closely to integer multiplication?

Mathematics

1 answer:

vitfil [10]3 years ago

5 0

I know the answer is

You might be interested in

The seventh multiple of 9 is<br>

weqwewe [10]

Answer:

First 7 multiples of 9 are:

9,18,27,36,45,54,63. = 36. Hope this helps :3

5 0

3 years ago

20 + 4y = 8<br> -21 2y = 1<br> Find the solution to the system of equations shown above by graphing,

Sonbull [250]

Note: Your system of equations is missing some details. Since the procedure is same so it should not be matter as it would still make your understand the concept. So I am assuming you have the following system of equations:

$20x\:+\:4y\:=\:8$

$-21x+\:2y\:=\:1$

Answer:

The two lines intersect at (3/31, 47/31) which is the solution to this system of equations.

The graph of the solution is also attached below.

Step-by-step explanation:

Given the system of the equations

$20x\:+\:4y\:=\:8$

$-21x+\:2y\:=\:1$

The solution will be the point of intersection of two lines. So we need to solve the system of the equations to find the point of intersection in order to graph the solution.

Solving

$\begin{bmatrix}20x+4y=8\\ -21x+2y=1\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:20x+4y=8:\quad x=\frac{2-y}{5}$

$\mathrm{Subsititute\:}x=\frac{2-y}{5}$

$\begin{bmatrix}-21\cdot \frac{2-y}{5}+2y=1\end{bmatrix}$

$\mathrm{Isolate}\:y\:\mathrm{for}\:-21\frac{2-y}{5}+2y=1$

$-21\cdot \frac{2-y}{5}+2y=1$

$-\frac{42}{5}+\frac{31y}{5}=1$ ∵ $-21\cdot \frac{2-y}{5}+2y= -\frac{42}{5}+\frac{31y}{5}$

$\mathrm{Multiply\:both\:sides\:by\:}5$

$-\frac{42}{5}\cdot \:5+\frac{31y}{5}\cdot \:5=1\cdot \:5$

$-42+31y=5$

$31y=47$

$\mathrm{Divide\:both\:sides\:by\:}31$

$\frac{31y}{31}=\frac{47}{31}$

$y=\frac{47}{31}$

$\mathrm{For\:}x=\frac{2-y}{5}$

$\mathrm{Subsititute\:}y=\frac{47}{31}$

$x=\frac{2-\frac{47}{31}}{5}$

$\frac{2-\frac{47}{31}}{5}=\frac{3}{31}$

$x=\frac{3}{31}$

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$y=\frac{47}{31},\:x=\frac{3}{31}$

Therefore, the two lines intersect at (3/31, 47/31) which is the solution to this system of equations.

The graph of the solution is also attached below.

7 0

3 years ago

In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win

icang [17]

There are $\binom{52}3=\frac{52!}{3!(52-3)!}$ (or "52 choose 3") ways of drawing any 3 cards from the deck.

There are 13 hearts in the deck, and 26 cards with a black suit. So there are $\binom{13}3$ and $\binom{26}3$ ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is

$P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}$

and the probability of drawing 3 black cards is

$P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}$

All other combinations can be drawn with probability $1-\frac{11}{850}-\frac2{17}=\frac{739}{850}$ .

Let $W$ be the random variable for one's potential winnings from playing the game. Then

$P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}$

a. For a single game, one can expect to win

$E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3$

b. For a single game, one's winnings have a variance of

$V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2$

where

$E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41$

so that $V[W]=\$^2\frac{1197}{17}\approx\$^270.41$ . (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.

c. With a $5 buy-in, the expected value of the game would be

$E[W-\$5]=E[W]-\$5=-\$2$

i.e. a player can expect to lose $2 by playing the game (on average).

d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:

$V[W-\$5]=V[W]$

so the standard deviation is the same, roughly $8.39.

e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.

8 0

3 years ago

Find the measure of angle E

Bad White [126]

Answer:

E = 76°

Step-by-step explanation:

Exterior angle B = 116°

So ABC = 180° - 116°

= 64°

In triangle ABC , AC = BC ( given )

So angle BAC = angle ABC = 64° ( Opposite angle of two same sides of a triangle are equal )

According to Angle Sum Property of a Triangle ,

BAC + ABC + ACB = 180°

=> ACB + 64° + 64° = 180°

=> ACB = 180° - 128°

= 52°

Also , angle ACB = angle ECD = 52° ( Vertical Opposite Angles are equal )

In triangle ECD ,

EC = ED ( given )

So, angle ECD = angle EDC = 52° ( Opposite angle of two same sides of a triangle are equal )

According to Angle Sum Property of a Triangle ,

ECD + EDC + CED = 180°

=> 52° + 52° + CED = 180°

=> CED = 180° - 104°

= 76°

5 0

3 years ago

Martha invested a principal amount of $15,000 into a savings account that earns simple interest at a rate of 4.5% per year. How

soldi70 [24.7K]

I=prt
I=15,000×0.045×6
I=4,050

3 0

3 years ago