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tekilochka [14]
3 years ago
7

Question 1 Find the absolute value of each integer. a. 141 = b. l-12] = c./-212=

Mathematics
1 answer:
Vilka [71]3 years ago
5 0

Answer:

is it a test and what is the lesson name called?

Step-by-step explanation:

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Kaitlin, Hans, Eric went a total of 98 Text messages during the weekend. Kaitlin 68 more messages than Hans, Eric sent three tim
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Answer:

Hans:6

Eric:18

Kaitlin:74

Step-by-step explanation:

first create an equation using the information in the question

5h+68=98

first subtract  68

5h+68=98

    -68  -68

5h=30

divide by 5

Hans:6

multiply by 3

Eric:18

68+6=74

Kaitlin:74

Hope this helps! Plz award Brainliest : )

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3 years ago
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Morgarella [4.7K]
81/200 is the answer
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Which statement about the hyperbola is true?
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2 years ago
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The points (–2, 4) and (–2, –2) are vertices of a heptagon. Explain how to find the length of the segment formed by these endpoi
Scrat [10]

Answer:

The segment is 6 units long.

Step-by-step explanation:

The points (–2, 4) and (–2, –2) are vertices of a heptagon. We have to explain how to find the length of the segment formed by these endpoints.

If two points at the ends of a straight line PQ are P(x_{1},y_{1}) and Q(x_{2},y_{2}), then the length of the segment PQ will be given by the formula  

\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}

Now, in our case the two points are (-2,4) and (-2,-2) and the length of the segment will be  

\sqrt{(- 2 - (- 2))^{2} + (4 - (- 2))^{2}} = 6 units. (Answer)

5 0
3 years ago
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I need help for this, thanks
gulaghasi [49]

(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.

---------------------------------------------------------------------------------------

(b) The following functions are positive in quadrant III:

tangent, cotangent

The following functions are negative in quadrant III

cosine, sine, secant, cosecant

A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.

5 0
3 years ago
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