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Serga [27]
3 years ago
8

5/2(6w-16)=18w-(3w+40) Pls help​

Mathematics
2 answers:
Dovator [93]3 years ago
6 0

5/2(6w - 16) = 18w - (3w + 40)

15w - 40 = 15w - 40

this equation is true.

valentinak56 [21]3 years ago
3 0

This is your answer <( ̄︶ ̄)↗

Step-by-step explanation:

.

.

#hope it helps you !!

(◕ᴗ◕)

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|45.28502&#10;-45.28|=0.00502,\quad |45.28502&#10;-45.29|0.00498

So, the distance from 45.29 is less than the distance from 45.28, which makes 45-29 the best approximation.

b. Rounding to the nearest integer  means that we can't use decimal digits at all. By the same logic of case (a), we have to choose between 27 and 28: we have

|27.499-27|=0.499,\quad |&#10;27.499-28|=0.501

Which makes 27 the best approximation

c. Rounding to the nearest tenth  means that we can only use one decimal digit. Again, the logic is always the same: the number lies between 0.2 and 0.3, and we use the same test to check which is a better approximation:

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For the x-values 1,2,3, and so on the y values of a function form a geometric sequence that decreases in value. What type of fun
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Step-by-step explanation:

6 0
4 years ago
Will someone please help me
Rufina [12.5K]

Answer:

The filled table for each equation by using the exact values in the table is

10x+2y=56                                          

x                                           y

______________________    

0                                          28

\frac{56}{10}                                          0

________________________

8x+3y=49

x                                               y

_________________________

0                                             \frac{49}{3}

\frac{49}{8}                                           0

_________________________

Step-by-step explanation:

Given equations are 10x+2y=56 and 8x+3y=49

To fill the table for each equation by using the exact values in the table :

10x+2y=56

put x=0 in above equation we get

10(0)+2y=56

2y=56

y=\frac{56}{2}

y=28

Therefore (0,28)

put y=0 in the given equation 10x+2y=56 we get

10x+2(0)=56

10x=56

x=\frac{56}{10}

Therefore (\frac{56}{10},0)

10x+2y=56

x                                               y

_________________________

0                                             28

\frac{56}{10}                                              0

__________________________

For

8x+3y=49

put x=0 in above equation we get

8(0)+3y=49

3y=49

y=\frac{49}{3}

Therefore (0,\frac{49}{3})

put y=0 in the given equation 8x+3y=49 we get

8x+3(0)=49

8x=49

x=\frac{49}{8}

Therefore (\frac{49}{8},0)

8x+3y=49

x                                               y

_________________________

0                                             \frac{49}{3}

\frac{49}{8}                                             0

________________________

7 0
4 years ago
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