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Gre4nikov [31]
3 years ago
11

Drag each function to the correct location on the chart. Classify the functions as continuous and discontinuous functions. If a

function is discontinuous, categorize it based on the type of discontinuity it exhibits.
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0

Answer:

Continuous: g(x) and j(x)

Removable: h(x) and m(x)

Infinite: f(x) and i(x) and k(x)

Jump: l(x)

Step-by-step explanation:

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James is working out and has decided that for every 10 sit-ups he does, he needs to do 4 push-ups. If he does 120 sit-ups, how m
mario62 [17]

Answer:

480 is the answer

Step-by-step explanation:

120x4 is 480

4 0
3 years ago
Quadrilateral PQRS is an isosceles trapezoid in which m∠P = m∠S. Which of the following statements must be true?
trasher [3.6K]

Answer:

letter B

Step-by-step explanation:

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it's correct!

7 0
3 years ago
PLEASE HELP!! please!!
lukranit [14]

Answer: dang that looks hard, are you actually gonna do that

Step-by-step explanation:

6 0
3 years ago
Boat A is 20 km from port on a bearing of 025° and boat B is 25 km from port on a bearing of 070°. Boat B is in distress. What b
tiny-mole [99]

Answer: Boat A should travel 152.8° to reach B

Step-by-step explanation:

The diagram illustrating the scenario is shown in the attached photo. Triangle ABP is formed. A represents the position of boat A. B represents the position of boat B. P represents the position of the port.

We would determine AB by applying the law of cosines

AB² = AP² + BP² - 2AP×BPCosP

AB² = 20² + 25² - 2 × 20 × 25 × Cos45

AB² = 1025 - 707.10678 = 317.89322

AB = √317.89322 = 17.83

We would determine the bearing of B from A by finding angle A. We would apply the sine rule.

AB/SinP = AP/Sin A

17.83/Sin45 = 20/SinA

Cross multiplying, it becomes

17.83 × SinA = 20Sin45 = 14.14

SinA = 14.14/17.83 = 0.79

A = Sin^-1(0.79) = 52.2°

The total angle at A is 65 + 52.2 = 117.2°

The angle formed outside the third quadrant is 117.2 - 90 = 27.2°

Therefore, bearing B from A is

180 - 27/2 = 152.8°

3 0
3 years ago
Find a gradient of a line that is parallel and perpendicular to this line with this gradient of -2
yulyashka [42]

a gradient of a line that is parallel and perpendicular to this line with this gradient of -2

Gradient is the slope

So slope of the line =-2

Slope of parallel line is equal to the slope of the line

So slope of parallel line = -2

Slope of perpendicular line is equal to negative reciprocal of slope of the line

We know slope of line = -2

Negative reciprocal = \frac{1}{2}

So , Slope of perpendicular line= \frac{1}{2}

7 0
3 years ago
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