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Masja [62]
4 years ago
5

ASAP!!! The distance (in units) between (−6, 2) and (8, 2).

Mathematics
1 answer:
Nesterboy [21]4 years ago
6 0

Answer:

14 units

Step-by-step explanation:

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Answer:

vertical asymptote X=2

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Dierdre’s work to solve a math problem is shown below.
kramer

Step-by-step explanation:

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Solve for the indicated variable. Include all of your work in your answer. Submit your solution.
kenny6666 [7]
C=2r
divide both sides by 2
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3 years ago
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HELP ASAP PLEASE!
djverab [1.8K]

Answer:

Step-by-step explanation:

The area of each rectangle is Length x width

Therefore Area of rectagle is 8cm x 10 cm=

A= 80 cm 2

There are three rectangles with the same measure so multiply the area by 3

80cm squared x 3= 240 cm 2

Now find the are of the triangle:

A- 1/2 x base x height

Base = 8cm

Height = 6.9 cm

A= (1/2) (8cm) (6.9cm)

A= (1/2) (55.2 cm squared)

A= 27.6 cm squared

Since there are two triangle faces with the same dimensions take the area and multiply it by 2

27.6 cm squared times 2 = 55/2 cm squared

Now add up the area of the rectangles and the area of the triangles.

Rectangles= 240 cm squared

Triangles 55.2 cm squared

240cm sq + 55.2 cm sq.= 295.2 cm squared

4 0
3 years ago
Find, correct to the nearest degree, the three angles of the triangle with the given ven
Scrat [10]

9514 1404 393

Answer:

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

Step-by-step explanation:

This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

  AB² = (4-1)² +(-3-0)² +(0-(-1)) = 16 +9 +1 = 26

  BC² = (1-4)² +(2-(-3))³ +(3-0)² = 9 +25 +9 = 43

  CA² = (1-1)² +(0-2)² +(-1-3)² = 4 +16 = 20

From the law of cosines, ...

  ∠A = arccos((AB² +CA² -BC²)/(2·AB·CA)) = arccos((26 +20 -43)/(2√(26·20)))

  ∠A = arccos(3/(4√130)) ≈ 86°

  ∠B = arccos((AB² +BC² -AC²)/(2·AB·BC)) = arccos((26 +43 -20)/(2√(26·43)))

  ∠B = arccos(49/(2√1118)) ≈ 43°

  ∠C = arccos((BC² +CA² -AB²)/(2·BC·CA)) = arccos((43 +20 -26)/(2√(43·20)))

  ∠C = arccos(37/(4√215)) ≈ 51°

The three angles are ...

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

_____

<em>Additional comment</em>

This sort of repetitive arithmetic is nicely done by a spreadsheet.

6 0
3 years ago
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