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leonid [27]
2 years ago
12

Questions in the attachments I’ll give a brainless

Mathematics
1 answer:
natulia [17]2 years ago
5 0

Answer:

Step-by-step explanation:

this has bean simplifyed dow to

= -1/6

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A regular octagon has side lengths of 8 centimeters. What is the approximate area of the octagon?
lara31 [8.8K]

Answer:

The answer is B if the octagon has side lengths of 8

5 0
1 year ago
If x = 4, then 5x = 20
Alexandra [31]

Answer:

Step-by-step explanation:

3 0
3 years ago
Solve for x 3(x+2)+4=5x+7
GaryK [48]
Distribute: 3x+6+4=5x+7
Simplify: 3x+10=5x+7
Subtract 3x from both sides: 10=2x + 7
Subtract 7 from both sides: 3=2x
Divide by 2: x=3/2

x=3/2
3 0
2 years ago
The yearly sales of candles decreased from 560 to 476. By what percentage did the candle sales decrease?
Korvikt [17]

The candle sales decreases to - 15 percentage.

<u>Step-by-step explanation:</u>

(476 - 560) : 560 x 100

= (476 : 560 - 1)100

= 85 - 100 = - 15

Therefore, - 15% did the candle sales decrease.

4 0
2 years ago
Jack works as a waiter and is keeping track of the tips he earns daily. about how much does jack have to earn in tips on sunday
OLga [1]

A mean is an arithmetic average of a set of observations. The amount that Jack has to earn in tips on Sunday if he wants to average $19 a day is $25.

<h3>What is Mean?</h3>

A mean is an arithmetic average of a set of observations. it is given by the formula,

\rm Mean=\dfrac{\text{Sum of all obervation}}{\text{Number of observation}}

As it is given that Jack makes $12 on Monday, $14 on Tuesday, $18 on Wednesday, $16 on Thursday, $26 on Friday and $22 on Saturday. And we need to know how much he should earn on Sunday, so the average is $19. Therefore, we can write,

\rm Mean=\dfrac{\text{Sum of all obervation}}{\text{Number of observation}}

19 = \dfrac{12+14+18+16+26+22+x}{7}\\\\133 = 108 + x\\\\133-108=x\\\\x = 25

Hence, the amount that Jack has to earn in tips on Sunday if he wants to average $19 a day is $25.

Learn more about Mean:

brainly.com/question/16967035

5 0
2 years ago
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