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3 years ago

How do you sovle this 5(x – 4) + 3x – 9x = 6 – (2x + 5) + 8x

Mathematics

1 answer:

sergeinik [125]3 years ago

3 0

Answer: <u>X=-3</u>

Step 1) Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

5*(x-4)+3*x-9*x-(6-(2*x+5)+8*x)=0

Step 2) ((5 • (x - 4) + 3x) - 9x) - (6x + 1) = 0

Step 3) Pull out like factors : -7x - 21 = -7 • (x + 3)

Step 4) -7 • (x + 3) = 0

Step 5) Solve : -7 = 0 This equation has no solution.

A a non-zero constant never equals zero.

Step 6) Solve : x+3 = 0

Subtract 3 from both sides of the equation :

x = -3

Step-by-step explanation:

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What decimal is smaller .52 or .25

Liono4ka [1.6K]

Answer:

0.25

Step-by-step explanation:

0.52 is like 2 quarters

0.25 is 1 quarter

So, 1 < 2

Have a nice day! :)

3 0

3 years ago

Due today please help!!!

Roman55 [17]

Answer:

y=(x+3)^2 -1

It goes left 3, so it is positive, for a mathematical reason. But then down one, so negative 1.

The subject is shifting parabolas. Tutorials online

3 0

3 years ago

What is the solution of x2+49 = x+5?

nikklg [1K]

Answer:

x = 12/5

Step-by-step explanation:

take the square of both sides

(x^2+49) = (x+5)^2

x^2 + 49 = (x+5) (x+5)

x^2+49= x^2 + 10x + 25

10x + 25 = 49

10x = 24

x = 12/5

4 0

3 years ago

A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p

andreyandreev [35.5K]

Answer:

B) 0.894

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.

A local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that $\mu = 0.55$

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577$

$P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894$

5 0

4 years ago

8 kg of mixed nuts containing 26% peanuts were mixed with 5 kg of another kind of mixed nuts that contain 52% peanuts. peanuts a

Gnesinka [82]

Answer:

Step-by-step explanation:

"8 kg of mixed nuts containing 26% peanuts"

weight of peanuts: 26% * 8 kg = 0.26 * 8 kg = 2.08 kg

"5 kg ... that contain 52% peanuts"

weight of peanuts: 52% * 5 kg = 0.52 * 5 kg = 2.6 kg

Total weight of peanuts: 2.08 kg + 2.6 kg = 4.68 kg

Total weight of mixed nuts: 8 kg + 5 kg

percent of peanuts: (4.68 kg)/(13 kg) * 100% = 36%

6 0

3 years ago