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2 years ago

Simplify

1 answer:

3 0

The first is a. you can rewrite it
$\sqrt{x^5} = \sqrt{x^2x^2x} = \sqrt{x^2}\sqrt{x^2}\sqrt{x} = xx\sqrt{x} = x^2\sqrt{x}$

the second is d. the only common factor between 20 and 40 that can be taken out is the 4 which becomes a 2, but inside the square roots there is still a 5 and 10, which cannot be combined

the third is c.
$\sqrt{36x} + \sqrt{100x} - \sqrt{25x} \\ 6\sqrt{x}+10\sqrt{x}-5\sqrt{x} \\ 11\sqrt{x}$

the fourth is d.
add 3 to both sides $\sqrt{2x} = 10$
then square both sides $2x=100$
then divide both sides by 2, x=50

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8_murik_8 [283]

Huh what about the photo you gotta b

6 0

2 years ago

a game consists of randomly choosing a bag (labelled 1, 2 or 3) out of a choice of 100, and then again randomly, picking a ball

Anton [14]

Answer:

hello your question has some missing information attached below is the missing information

answer : 112 black balls

Step-by-step explanation:

Given data:

Each bag contains ; 10 balls

Total number of ball = 100 balls

first: calculate the number of Black balls to be picked from the three bags

From bag 1 = 0.4 * 0.35 * 10 = 1.4 balls

From bag 2 = 0.8 * 0.45 * 10 = 3.6 balls

From bag 3 = 0.3 * 0.2 * 10 = 0.6 ball

Total black balls picked = 5.6

∴ percentage of black balls picked given that each bag contains 10

5.6 / 10 * Total balls available = 56 balls = 56% of 100 balls

Hence If the player plays this game 200 times the number of black balls he will pick = 56% * 200 = 112 balls

4 0

2 years ago

HELP PLEASE 10 POINTS!!!

7nadin3 [17]

Answer:

First, write 80% as 80100. Then write an equivalent ratio that has the answer in the numerator and 200 in the denominator. 100 x 2 is 200, so 80 x 2 will be the answer. 80 x 2 = 160. The answer is $160.

3 0

2 years ago

In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose

viktelen [127]

Answer:

3 students

Step-by-step explanation:

If everyone in the class has either a pierced nose or ear, we just simply have to add up the total number of hands raised and minus the number of students in the class.

35+8=43

43-40=3

3 students have both a pierced nose and pierced ear.

6 0

2 years ago

What is the center of the circle that you can circumscribe about a triangle with vertices A(2, 6), B(2, 0), and C(10, 0)?

denis23 [38]

The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle (x,y):
$d^2 = (x-x_0)^2 + (y-y_0)^2$
Plugging in points A and B into distance formula, then setting them equal to each other gives:
$(x-2)^2+(y-6)^2 = (x-2)^2 + y^2$
Right away we can cancel out the x terms leaving:
$(y-6)^2 = y^2$
Expand Left side and Solve for y:
$y^2 -12y +36 = y^2$
$y = 3$
Plug in points B and C as before:
$(x-2)^2 + y^2 = (x-10)^2 + y^2$
Here we can cancel the y-terms.
Expand and solve for x:
$x^2 -4x+4 = x^2 -20x+100$
$16x = 96$
$x = 6$
Therefore the center of the circle is the point (6,3)

6 0

3 years ago