Solve for this linear equation xm=np, solve for x

X^2+6x+9=1 Solve each equation.

devlian [24]

It is -4 and -2 for X

6 0

4 years ago

Read 2 more answers

I need to find the length of MK.

Morgarella [4.7K]

Answer:

Here,

JM=MK

so,

7x+5 = 8x

$7x + 5 = 8x \\ ⇒ 5 = 8x - 7x \\ ⇒5 = x$

. ° . JM = 7x+5 = 7×5+5 = 40

. ° . MK = 8x = 8×5 = 40

Hence,

5 0

3 years ago

Is 5 1/4 / 3 1/2 the same as 3 1/2 / 5 1/4? Explain.

yan [13]

No it is not the same because you can not reverse division problems the same way you can with multiplication.

5 1/4 / 3 1/2 is asking how many times 3 1/2 can go into 5 1/4. The answer to this would be 1.5

3 1/2 / 5 1/4 is asking how many times 5 1/4 can go into 3 1/2. The answer to this would be .667

3 0

4 years ago

Use the Counting Principle to find the probability of rolling a 4 on each of 3 number cubes

Dennis_Churaev [7]

Answer:

$\frac{1}{216}$

Step-by-step explanation:

We have been given that three cubes are being rolled and we are required to find the probability of getting 4 on each of the three cubes.

We know the probability of rolling a 4 on one cube will be $\frac{1}{6}$ . Since there are 3 cubes, the probability of rolling 4 in each of the individual cubes will be $\frac{1}{6}$ .

But we want 4 to be rolled on all three cubes at once, in order for that to happen, we need to get a four on first cube AND second cube AND third cube. Since there is AND between the events, we will multiplied their probabilities as per counting principle. Thus, the required probability of rolling 4 on each of the three cubes is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}$ which simplifies to $\frac{1}{216}$ .

8 0

3 years ago

If points A and B are randomly placed on the circumference of a circle with a radius of 2 cm, what is the probability that the l

OLga [1]

Answer:

The probability that the length of chord AB > 2 cm is 2/3 ⇒ D

Step-by-step explanation:

* Lets explain how to solve the problem

- Points A and B are randomly placed on the circumference of a

circle with a radius of 2 cm

- We need to find the probability that the length of chord AB is

greater than 2 cm

- Probability = required length of arc / circumference

- Assume that the length of the chord AB is 2 and the center of

the circle is labeled M

- The radius of the circle is 2 and AB = 2

- In Δ AMB

∵ MA= MB = 2 radii

∵ AB = 2

∴ Δ AMB is an equilateral triangle

∴ m∠AMB = π/3

∵ The length of an arc = r Ф , where r is the radius of the circle and

Ф is the central angle subtended by this arc

∵ r = 2 , Ф = π/3

∴ The length of arc AB = 2 × π/3 = 2π/3

- The arc AB is in the half of the circle then there are another arc

with the same length in the other half of the circle

- The length of the arc we must excluded from the length of the

circle to have the part of length of AB is greater than 2 is

2(2π/3) = 4π/3

∴ The length of the excluded arc is 4π/3

∵ The length of the circle is 2πr

∵ r = 2

∴ The length of the circle = 2π(2) = 4π

- The required arc = length circle - length excluded arc

∵ The length of the excluded arc = 4π/3

∵ The length of the circle = 4π

∴ The required arc = 4π - 4π/3 = 8π/3

- Probability = required length of arc / circumference

∴ Probability = (8π/3)/(4π) = 2/3

∴ The probability that the length of chord AB > 2 cm is 2/3

4 0

3 years ago