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3 years ago

Factoriza completamente las siguientes expresiones;

1 answer:

8 0

<span>1. 3a²-a = a(3a - 1)

2. 7ab³-14b = 7b(ab^2 - 2)

3. x³-x²+x = x(x^2 - x + 1)

4. 12x³-xy² = x(12x^2 - y^2)

5. 5x-6x²+7x³ = x(5 - 6x + 7x^2)

6. 7a+8ab+9a² = a(7 + 8b + 9a)

7.3x²-3xy+6xy² </span>= 3x(x - y + 2y^2)

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A bag contains 10 red marbles, 15 yellow marbles, 5 green marbles, and 20 blue marbles. Five marbles are drawn from the bag. wha

Charra [1.4K]

To determine the probability that exactly two of the five marbles are blue, we will use the rule of multiplication.

Let event A = the event that the first marble drawn is blue; and let B = the event that the second marble drawn is blue.
To start, it is given that there are 50 marbles, 20 of them are blue. Therefore, P(A) = 20/50
After the first selection, there are 49 marbles left, 19 of them are blue. Therefore, P(A|B) = 19/49
Based on the rule of multiplication:P(A ∩ B) = P(A)*P(A|B)P(A ∩ B) = (20/50) (19/49)P(A ∩ B) = 380/2450P(A ∩ B) = 38/245 or 15.51%

The probability that there will be two blue marbles among the five drawn marbles is 38/245 or 15.51%

We got the 15.51% by dividing 38 by 245. The quotient will be 0.1551. We then multiplied it by 100% resulting to 15.51%

7 0

3 years ago

LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

5 0

2 years ago

What is another name for the height of the pyramid?

Lynna [10]

The altitude might be another name for a height of a pyramid, lol, hope this helps

5 0

3 years ago

8x - 11 = -2x + 15 -5. What steps would you take to solve this?

MAVERICK [17]

Combine like terms. So that means combine the X's and combine the regular numbers.

- add 2x to 8x
- combine +15 -5
- then add 11 to that
- once you're done, find X by dividing 10x by whatever you have after the = sign

8 0

2 years ago

Which of the following has the same value as

Gwar [14]

Answer:

i did this 2 months ago, and i believe its A if not sorry

Step-by-step explanation:

3 0

1 year ago

Read 2 more answers