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3 years ago

Simplify (right answer only ) : 2(x – 9)

Mathematics

2 answers:

Ann [662]3 years ago

6 0

Answer:

2x-18

Step-by-step explanation:

2(x-9)

distribute the 2 in the ( )

2x-18

Hope this Helps!!!

skad [1K]3 years ago

3 0

2x-18 because you distribute the x so you multiply the 2 with x which is 2x and then the 2*9 is 18

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The bill at the end of a meal was £20.40 .10% was added as a service charge how much wad added

Dmitrij [34]

Answer:

$2.04

Step-by-step explanation:

The bill at the end of a meal was £20.40

10% was added

service charge was added = $(20.40 x 10/100) [x%=x/100]

=$2.04

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3 years ago

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At the embryonic stage, the zygote becomes the embryo. T or F

ehidna [41]

Answer:

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4 years ago

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A pathologist has been studying the frequency of bacterial colonies within the field of a microscope using samples of throat cul

nata0808 [166]

Answer:

P(0) = 0.055

P(1) = 0.16

P(2) = 0.231

P(3) = 0.224

P(4) = 0.162

P(5 or more) = 0.168

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Here λ = 2.90 is the average number of bacteria colonies per field.

This means that $\mu = 2.90$

Compute P(r) for r = 0, 1, 2, 3, 4, and 5 or more.

$P(0) = \frac{e^{-2.9}*(2.9)^{0}}{(0)!} = 0.055$

$P(1) = \frac{e^{-2.9}*(2.9)^{1}}{(1)!} = 0.16$

$P(2) = \frac{e^{-2.9}*(2.9)^{2}}{(2)!} = 0.231$

$P(3) = \frac{e^{-2.9}*(2.9)^{3}}{(3)!} = 0.224$

$P(4) = \frac{e^{-2.9}*(2.9)^{4}}{(4)!} = 0.162$

5 or more:

This is

$P(X \geq 5) - 1 - P(X < 5)$

In which:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.055 + 0.16 + 0.231 + 0.224 + 0.162 = 0.832$

$P(X \geq 5) - 1 - P(X < 5) = 1 - 0.832 = 0.168$

P(5 or more) = 0.168

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3 years ago

The greatest common factor of 36 and 45

elena55 [62]

I believe the answer is 9

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4 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

8 0

4 years ago