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3 years ago

Simplify the following radical expression : 3√12 × 5√6

1 answer:

7 0

$3 \sqrt{12} *5 \sqrt{6} = 15 \sqrt{72}$ .

Simplifying the radical, we have $15 \sqrt{72} = 15*6 \sqrt{\frac{72}{6^{2}} } =90 \sqrt{2}$ , so our answer is $90 \sqrt{2}$ .

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Determine the coordinates of the point on the unit circle corresponding to the given central angle. If necessary round your resu

Olenka [21]

Answer:

Option A. (-1, 0)

Step-by-step explanation:

In the figure attached,

Circle O is a unit circle (having radius r = 1 unit)

If a point A with central angles = θ, is lying on the circle then the coordinates of the point A will be,

x = r.cosθ

x = 1.cosθ = cosθ

and y = r.sinθ

y = 1.sinθ = sinθ

Therefore, coordinates representing the point A will be (cosθ, sinθ).

As per question the given point A is lying at P (a point having central angle θ = 180°)

Coordinates of point P will be

(x', y') → (cos180°, sin180°)

→ (-1, 0)

Therefore, Option A will be the answer.

3 0

3 years ago

The area of the shaded sector is shown. Find the radius of circle M. Round your answer to the nearest hundredth.

Vaselesa [24]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the sector, its area and the angle at the center are not given.

I will solve this using the following illusration

The area of a sector is:

$Area= \frac{\theta}{360} * \pi r^2$

Assume that:

$\theta = 120$

$Area = 20$

The equation becomes

$20= \frac{120}{360} * \pi r^2$

Simplify

$20= \frac{1}{3} * \pi r^2$

Take: $\pi = 3.14$

$20= \frac{1}{3} *3.14* r^2$

Cross Multiply

$3.14 * r^2 = 20 * 3\\$

Solve for $r^2$

$r^2 = \frac{20 * 3}{3.14}$

$r^2 = 19.11$

Take square roots

$r = \sqrt{19.11$

$r = 4.4$

3 0

3 years ago

How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com

Mnenie [13.5K]

If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)

(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.

If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)

(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.

We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845

Therefore, there are 4845 ways to have a group of 4 with 4 women.

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee

5 0

3 years ago

Solve the quadratic by quadratic formula 7x-10=x squared

alexgriva [62]

0= x^2-7x+10

Factor
(x-5)(x-2)=0

Check
-5-2=-7, -5*-2=10

Use ZPP
x-5=0
x=5

x-2=0
x=2

Final answer: x=2, x=5

4 0

3 years ago

4 2/3 + 1 3/4 as a fraction

Alex787 [66]

Answer:

77/12

Step-by-step explanation:

sorry is it's wrong

4 0

2 years ago