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2 years ago

Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15.

Mathematics

2 answers:

Troyanec [42]2 years ago

6 0

Answer:

Both absolute values would need to be isolated first.

You would need to write a compound inequality for each.

Both compound inequalities would compare

x – 1 to –15 and 15.

The inequality with “<” would use an “and” statement, while the “>” would use an “or” statement.Step-by-step explanation:

kicyunya [14]2 years ago

4 0

In the first case we'd subtract 1 from both sides, obtaining |x-1|<14.

In the second case we'd also subtract 1 from both sides, and would obtain
|x-1|>14.

What would the graphs look like?

In the first case, the graph would be on the x-axis with "center" at x=1. From this center count 14 units to the right, and then place a circle around that location (which would be at x=15). Next, count 14 units to the left of this center, and place a circle around that location (which would be -13). Draw a line segment connecting the two circles. Notice that all of the solutions are between -13 and +15, not including these endpoints.

In the second case, x has to be greater than 15 or less than -13. Draw an arrow from x=1 to the left, and then draw a separate arrow from 15 to the right. None of the values in between are solutions.

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Determine the value of (tangent of 88 degrees, 22 minutes, and 45 seconds). In this case, minutes are 1/60 of a degree and secon

mariarad [96]

Answer:

c. 35.34015106

Step-by-step explanation:

As with many problems of this nature, you only need to get close to be able to choose the correct answer. 22 minutes 45 seconds is just slightly less than 1/2 degree (30 minutes), so the tangent value will be just slightly less than tan(88.5°) ≈ 38. The appropriate choice is 35.34015106.

If you need confirmation, you can find tan(88°) ≈ 29, so you know the answer will be between 29 and 38.

The above has to do with strategies for choosing answers on multiple-choice problems. Below, we will work the problem.

The angle is (in degrees) ...

88 + 22/60 +45/3600 = 88 + (22·60 +45)/3600 = 88 +1365/3600

≈ 88.3791666... (repeating) . . . . degrees

A calculator tells you the tangent of that is ...

tan(88.3791666...°) ≈ 35.3401510614

Many calculators will round that to 10 digits, as in the answer above. Others can give a value correct to 32 digits. Spreadsheet values will often be correct to 15 or 16 digits.

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2 years ago

Suppose that we ask n randomly selected people whether they share your birthday. (a) Give an expression in terms of n for the pr

expeople1 [14]

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1 year ago

Titus is asked to prove hexagon FEDCBA is congruent to hexagon

Goshia [24]

Answer:

(x, y) -------> ( - x , y - 10 )

Step-by-step explanation:

Titus is not correct.

There are two transformations will correctly prove FEDCBA ≅ F'E'D'C'B'A'

First transformation is Reflection over y-axis , then translation 10 units down.

The final formula is ( - x , y - 10 )

5 0

1 year ago

At the beginning of the business day, a bank's vault held $575,900. By the end of the day, $(3.5 103) had been added to the vaul

Likurg_2 [28]

3.5*10^3=3,500

3,500+ 575,900=579,400

5 0

2 years ago

The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% R

IRISSAK [1]

Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

Before 1995

(Br) Brown = 30%

(Y) Yellow = 20% =0.2

(R) Red = 20%

(G) Green =10% =0.1

(O) Orange = 10%

(T) Tan = 10%

After 1995

(Br) Brown = 13%

(Y) Yellow = 14% =0.14

(R) Red = 13%

(G) Green = 20% = 0.2

(O) Orange = 16%

(Bl) Blue = 24%

Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

AG implies we drew a green M&M from the 1994 bag

BY implies imply we drew a yellow M&M from the 1996 bag

BG implies we drew a green M&M from the 1996 bag

P(AY) =0.2

P (BY) = 0.14

P(AG) =0.1

P(BG) =0.2

Since the draws from the 1994 and 1996 bag are independent,

therefore

$P(AY n BG) = 0.2 * 0.2 = 0.04 -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014 --------(2)\\$

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E = $P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54$

For the yellow one to be from 1994, it implies that the event to be chosen is

$P(AYnBG) = 0.2*0.2$

Since the total probability is given as E=0.054

then $P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407$

Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E

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2 years ago