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d1i1m1o1n [39]
3 years ago
11

Coordinate grid with graph of a straight line with a negative slope. The line passes through the first, second, and fourth quadr

ants. Evaluate the function at x = –3. A. y = 4 B. y = 3 C. y = –3 D. y = 0
Mathematics
1 answer:
Blababa [14]3 years ago
4 0
Since the straight line has a negative slope and passes therough the first, second and fourth quadrants, the value of y when x = -3 will be a positive number greater than 3.

Therefore, the correct answer is option A (y = 4)
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26.72 ÷ 4.7 (Round to the nearest hundredth)
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6 0r 5.6 rounded to the nearest hundredth

Step-by-step explanation:

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A(n) = -5 + 6(n - 1)
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domain is all reall numbers

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ASAP!!! Use the pythagorean theorem to prove that the point (√2/2, √2/2) lies on the unit circle. I need setup, explination, ans
docker41 [41]

Answer:

In brief, apply the pythagorean theorem to show that the distance between the point (\sqrt{2}/2,\sqrt{2}/2) and the origin is 1.

Step-by-step explanation:

The pythagorean theorem can give the distance between two points on a plane if their coordinates are known.

A point is on a circle if its distance from the center of the circle is the same as the radius of the circle.

On a cartesian plane, the unit circle is a circle  

  • centered at the origin (0,0)
  • with radius 1.

Therefore, to show that the point (\sqrt{2}/2,\sqrt{2}/2) is on the unit circle, show that the distance between (\sqrt{2}/2,\sqrt{2}/2) and (0,0) equals to 1.

What's the distance between (\sqrt{2}/2,\sqrt{2}/2) and (0,0)?

\displaystyle \sqrt{\left(\frac{\sqrt{2}}{2}-0}\right)^{2} + \left(\frac{\sqrt{2}}{2}-0\right)^{2}} = \sqrt{\frac{1}{2} + \frac{1}{2}}= \sqrt{1}= 1.

By the pythagorean theorem, the distance between (\sqrt{2}/2,\sqrt{2}/2) and the center of the unit circle, (0,0), is the same as the radius of the unit circle, 1. As a result, the point (\sqrt{2}/2,\sqrt{2}/2) is on the unit circle.

3 0
3 years ago
An edge of a square has vertices with coordinates (a, b) and (-a,- b). Write expre
Tema [17]

Answer:

Step-by-step explanation:

Value of the edge of the square=d

d^2=(a^2+b^2+(-a)^2+(-b)^2=2(a^2+b^2)\\\\d=\sqrt{2(a^2+b^2)}\\\\Perimeter=4*d=4*\sqrt{2(a^2+b^2)}\\\\Area=d^2=2(a^2+b^2)\\

6 0
3 years ago
Four students (A, B, C, and D) are interviewing for an all-expenses-paid vacation to a country of their choice. Only one student
Shkiper50 [21]

Answer:

A. 0.36 ; B. 0.72

Step-by-step explanation:

Probability of A, B, C, D to win the interview = P(A), P(B), P(C), P(D)

Given : P(A) = 2 P(B)    ∴ P(B) = P(A) / 2

P(B) = 2/3 P(C)   ∴ P(C) = 3/2 P(B)   ∴ P(C) = 3/2 [P(A) / 2 ]

So, P(C) = 3/4 P(A)

P(C) = 1.5 P(D)   ∴ P(C) = 3/2 P(D)      ∴ P(D) = 2/3 P(C)

∵ P(C) = 3/4 P(A)   ∴ P(D) = 2/3 [3/4 P(A)]

So, P(D) = 1/2 P(A)

Either of them will definitely win the interview. So probability of A or B or C or D winning = 1

So, P(A) + P(B) + P(C) + P(D) = 1

Putting above values : P(A) + P(A) / 2 + 3/4 P(A) + 1/2 P(A)  =  1

P(A) [1+ 1/2 + 3/4+ 1/2]  =   [(4+2+3+2) /4] P(A)

∴ 2.75 P(A) = 1

So, P(A) = 1/2.75 = 0.36

P(B) = P(A)/2 = 0.36 /2 = 0.18

P(C) = 3/4 P(A) = 0.36 (3/4) = 0.27

P(D) = P(A)/2 = 0.36 /2 = 0.18

A. Probability A wins election = 0.36

B. Probability C doesn't win election = Pr A or B or D win election

= Pr (A) + Pr (B) + Pr (D) = 0.36 + 0.18 + 0.18 = 0.72

4 0
3 years ago
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