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1 year ago

10a + 6b + 6cIt needs to be simplified I also need the steps

Mathematics

1 answer:

Harlamova29_29 [7]1 year ago

6 0

Since all the numbers that accompany the letters a, b and c are even, then you can apply common-factor to factor the 2.

Then you have

$\begin{gathered} 10a+6b+6c=2\cdot5a+2\cdot3b+2\cdot3c \\ 10a+6b+6c=2(5a+3b+3c) \end{gathered}$

Therefore, the given simplified expression is

$2(5a+3b+3c)$

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Simplify the expressions by distributing and combining like terms 5/8x-2(7-1/4x)

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9x/18 = 14 I’m not sure if I’m right but I think it’s that

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3 years ago

Find the length of the hypotenuse of a right triangle whose legs are 3 and . 2

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Answer:

The answer would be 3.606, but rounded is 3.61, or 3.6

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3 years ago

1) How much money will you have in an account after 5 years if you invest $1,200 at 2.5%

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Answer:

$ 1,350

Step-by-step explanation:

i = (prt) / 100

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3 years ago

3.14 The waiting time, in hours, between successive speeders spotted by a radar unit is a continuous random variable with cumula

vovangra [49]

Answer:

(a) The probability of waiting less than 12 minutes between successive speeders using the cumulative distribution function is 0.7981.

(b) The probability of waiting less than 12 minutes between successive speeders using the probability density function is 0.7981.

Step-by-step explanation:

The cumulative distribution function of the random variable <em>X, </em>the waiting time, in hours, between successive speeders spotted by a radar unit is:

$F(x)=\left \{ {{0;\ x$

(a)

Compute the probability of waiting less than 12 minutes between successive speeders using the cumulative distribution function as follows:

$12\ \text{minutes}=\frac{12}{60}=0.20\ \text{hours}$

The probability is:

$P(X$

$=(1-e^{-8x})|_{x=0.20}\\\\=1-e^{-8\times 0.20}\\\\=0.7981$

Thus, the probability of waiting less than 12 minutes between successive speeders using the cumulative distribution function is 0.7981.

(b)

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{d F (x)}{dx}=\left \{ {{0;\ x$

Compute the probability of waiting less than 12 minutes between successive speeders using the probability density function as follows:

$P(X$

$=8\times [\frac{-e^{-8x}}{8}]^{0.20}_{0}\\\\=[-e^{-8x}]^{0.20}_{0}\\\\=(-e^{-8\times 0.20})-(-e^{-8\times 0})\\\\=-0.2019+1\\\\=0.7981$

Thus, the probability of waiting less than 12 minutes between successive speeders using the probability density function is 0.7981.

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