All categories

2 years ago

Find the product.

1 answer:

3 0

Answer :

$6{y}^{4} {z}^{2} + 12 {y}^{3} {z}^{2}-3 {y}^{3} {z}+3 {y}^{2} {z}^{2}$

Step-by-step explanation :

To find the product of

$3 {y}^{2} z(2 {y}^{2} z + 4yz - y + z)$

First we expand the bracket ,

it implies that, we use the expression outside the bracket to multiply individual expressions inside the bracket.

Hence

$3 {y}^{2} z(2 {y}^{2} z + 4yz - y + z)$

$= 3 {y}^{2} z(2 {y}^{2} z) + 3 {y}^{2} z(4yz) - 3 {y}^{2} z(y )+3 {y}^{2} z( z)$

we now apply the law of indices

${a}^{m} \times {a}^{n} = {a}^{m+n}$

meaning, when you are multiplying two expressions with the same bases , repeat one of the bases and add the exponents.

Then, simplify to obtain

$= 6{y}^{4} {z}^{2} + 12 {y}^{3} {z}^{2}-3{y}^{3} {z}+3 {y}^{2} {z}^{2}$

You might be interested in

(6x-5)+(-7x+2) and (6x-5)-(-7x+2) Need work

tia_tia [17]

Answer:(6x-5)+(-7x+2)

1 Remove parentheses.

6x-5-7x+2

2 Collect like terms.

(6x-7x)+(-5+2)

3 Simplify.

-x-3

_____________________________

1 Remove parentheses.

6x-5+7x-2

6x−5+7x−2

2 Collect like terms.

(6x+7x)+(-5-2)

3 Simplify.

13x-7

Hope this helps

-Tobie the dog <3

7 0

2 years ago

What is the total surface area of the drawing?

kkurt [141]

The answer will be 21.2kilometers

7 0

3 years ago

Each day, X arrives at point A between 8:00 and 9:00 a.m., his times of arrival being uniformly distributed. Y arrives independe

astraxan [27]

Answer:

Y will arrive earlier than X one fourth of times.

Step-by-step explanation:

To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:

$f_X(x) = \frac{1}{L}$

Where L stands for the length of the interval over which the variable is distributed.

Now, as X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:

$f_X(x) = 1\\\\f_Y(y)=2$ .

Now, the probability of an event is equal to the integral of the density probability function:

$\iint_A f_{X,Y} (x,y) dx\, dy$

Where A is the in which the event happens, in this case, the region in which Y<X (Y arrives before X)

It's useful to draw a diagram here, I have attached one in which you can see the integration region.

You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.

Let's set up the integration:

$\iint_A f_{X,Y} (x,y) dx\, dy\\\\\iint_A f_{X} (x) \, f_{Y} (y) dx\, dy\\\\2 \iint_A dx\, dy$

We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:

$2 \iint_A dx\, dy= 2 \times \frac{0.5 \times 0.5 }{2} \\\\2 \iint_A dx\, dy= \frac{1}{4}$

That means that one in four times Y will arrive earlier than X. This result can also be seen clearly on the diagram, where we can see that the triangle is a fourth of the rectangle.