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polet [3.4K]
3 years ago
11

Hello i need help on thus​

Mathematics
2 answers:
marusya05 [52]3 years ago
6 0

Answer:

The answer is B

Step-by-step explanation:

Illusion [34]3 years ago
5 0

Answer:

I believe the answer is

B. \: 15, \: 23, \: 27, \: - 24, \:  - 9

Step-by-step explanation:

Hope thai helps!

please like if it does!

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Which situation is most likely to show a constant rate of change? A. The number of people on a city bus compared with the time o
Marina86 [1]
<span>D. The amount spent on grapes compared with the weight of the purchase. In the other examples, the rate of change would be either very erratic (people on the bus, which would change suddenly and in large amounts depending upon stops, the girl's shoe size, which would change somewhat consistently, but with swings as growth spurts happened, or not at all, like the pizza example, which if you think about it, the amount of pizza doesn't really matter). The grape answer would have a set amount of pay based on weight (say $1 per pound), which would have a constant rate of change tied to the weight.</span>
3 0
4 years ago
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There are 2 green,5 yellow, 6 red,and 7 blue marbles in a bag. Erin picks one marble from the bag without looking. What is the p
Svetach [21]

Answer:

c 35%

Step-by-step explanation:

2 green,5 yellow, 6 red,and 7 blue marbles

2+5+6+7 = 20 marbles

P(blue) = blue marbles / total marbles

             = 7/20

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7 0
3 years ago
Solve <br><br> 12 1/2-(-4 1/2)=
Masteriza [31]

12½ -(-4½)

= 24+1/2 + 8+1/2

= 25/2 + 9/2

= 34/2

= 17.

4 0
3 years ago
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Round 514,733 to the nearest thousand
sergij07 [2.7K]
515,000 is the nearest
3 0
3 years ago
Read 2 more answers
Pls show work/steps for both questions
aleksklad [387]

Answer:

Step-by-step explanation:

\frac{-5+2 \iota}{1+7 \iota}  \times \frac{1-7 \iota}{1-7 \iota} \\=\frac{-5+35 \iota+2 \iota-14 \iota^2}{1-49 \iota^2} \\=\frac{-5+37 \iota-14(-1)}{1-49(-1)} \\=\frac{9+37 \iota}{50} \\=\frac{9}{50} +\frac{37}{50}  \iota

2.

domain is x²≤36

|x|≤6

-6≤x≤6

or [-6,6]

3 0
3 years ago
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