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2 years ago

15

Find the greatest common factor of these three expressions.

1 answer:

leva [86]2 years ago

3 0

6x

42/6 = 7 (valid), 18/6 = 3 (valid), 12/6 = 2 (valid)

7, 8, 9, 10, and 11 are all not divisible by 12, and 12 is not divisible by 18, so 6 is the highest number.

Also, each number has at least one x, so you can take one out.

Remaining=
6x (7 + 3x^2 + 2x^3)

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Expand and simplify (x-3)(2x+1)(x+1)

Vladimir [108]

Answer:

2 3 − 3 2 − 8 − 3

Step-by-step explanation:

6 0

3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

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snow_lady [41]

8-1/4n is the answer to this

7 0

3 years ago

PLS HELP WILL GIVE BRAINIEST! You are flying in an airplane that has 2 engines. The engines individually have a success rate of

Salsk061 [2.6K]

Answer:

.01%*

Step-by-step explanation:

Individually, they both have a rate of failure of 1%.

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*Disclaimer!!

I am not the best at this area in math so I would still reference against someone else's answer, but this is my best effort :)

4 0

3 years ago

(04.02 MC)<br> Solve the system of equations.<br> 5x + y = 9<br> 3x + 2y = 4

goblinko [34]

Answer:

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Step-by-step explanation:

Multiply first equation by 2:

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Subtract the second equation from this so that the 2y terms cancel:

10x - 3x = 18 - 4

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x = 2

Plug into first equation to find y:

5(2) + y = 9

10 + y = 9

y = -1

The answer is (2,-1)

7 0

3 years ago

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