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Zielflug [23.3K]
3 years ago
12

Add like terms 5x + 2y + 2x - 4y - 3x

Mathematics
2 answers:
solong [7]3 years ago
4 0

Answer:

4x-2y

Step-by-step explanation:

Tju [1.3M]3 years ago
4 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{4x - 2y}}}}}

Step-by-step explanation:

\sf{5x + 2y + 2x - 4y - 3x}

Like terms are those which have the same base.

Combine the like terms and simplify

\dashrightarrow{ \sf{5x + 2x  - 3x + 2y - 4y}}

\dashrightarrow{ \sf{4x - 2y}}

Hope I helped!

Best regards! :D

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Answer:

The Answer is 400

Step-by-step explanation:

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There are 15 marbles in a bag. Three marbles are red, two marbles are blue, four marbles are green, and six marbles are yellow.
Vesnalui [34]

Answer:

6/15 is probability

Step-by-step explanation:

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4 years ago
The picture. and answer 1-6. please hurry
satela [25.4K]

Answer:

Step-by-step explanation:

From the figure attached,

1). Sin(D) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

               = \frac{EF}{DF}

               = \frac{24}{26}

               = \frac{12}{13}

2). Cos(D) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

                = \frac{ED}{DF}

                = \frac{10}{26}

                = \frac{5}{13}

3). tan(D) = \frac{\text{Opposite side}}{\text{Adjacent side}}

               = \frac{EF}{DE}

               = \frac{24}{10}

               = \frac{12}{5}

4). sin(F) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

              = \frac{DE}{DF}

              = \frac{10}{26}

              = \frac{5}{13}

5). cos(F) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

               = \frac{EF}{DF}

               = \frac{24}{26}

               = \frac{12}{13}

6). tan(F) = \frac{\text{Adjacent side}}{\text{Adjacent side}}

               = \frac{DE}{EF}

               = \frac{10}{24}

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4 0
3 years ago
−4 /5+1 /5 = <br> please help me
MaRussiya [10]

Answer:

-0.6

Step-by-step explanation:

think of it like this

(-4/5) +(1/5)

5 0
3 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
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