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2 years ago

Expand (2x-3y)^4 using Pascal's Triangle. Show work

1 answer:

3 0

(2x+3y)⁴
1) let 2x = a and 3y = b

(a+b)⁴ = a⁴ + a³b + a²b² + ab³ + b⁴
Now let's find the coefficient of each factor using Pascal Triangle

0 | 1
1 | 1 1
2 | 1 2 1
3 | 1 3 3 1
4 | 1 4 6 4 1

0,1,2,3,4,.. represent the exponents of binomials
Since our binomial has a 4th exponents, the coefficients are respectively:

(1)a⁴ + (4)a³b + (6)a²b² + (4)ab³ + (1)b⁴
Now replace a and b by their real values in (1):

2⁴x⁴ +(4)8x³(3y) + (6)(2²x²)(3²y²) + (4)(2x)(3³y³) + (1)(3⁴)(y⁴)

16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴

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A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago

Given \qquad \overline{PQ}\perp\overline{PS} PQ ⊥ PS start overline, P, Q, end overline, \perp, start overline, P, S, end over

Tomtit [17]

Answer: 40 degrees

Step-by-step explanation:

did this on khan! :)

5 0

2 years ago

Which two equations would be most appropriately solved by using the zero product property? Select each correct answer.

LekaFEV [45]

The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.

For more context let's look at the first equation in the problem that we can apply this to: $(x-3)(x+4)=0$

Through zero property we know that the factor $(x-3)$ can be equal to zero as well as $(x+4)$ . This is because, even if only one of them is zero, the product will immediately be zero.

The zero product property is best applied to factorable quadratic equations in this case.

Another factorable equation would be $2x^{2}+6x=0$ since we can factor out $2x$ and end up with $2x(x+3)=0$ . Now we'll end up with two factors, $2x$ and $(x+3)$ , which we can apply the zero product property to.

The rest of the options are not factorable thus the zero product property won't apply to them.

3 0

2 years ago

Read 2 more answers

Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real

adoni [48]

Step-by-step explanation

<h3>Prerequisites:</h3>

$b^2 - 4ac = 0 \Rightarrow 1 solution$