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Ann [662]
3 years ago
7

The diagram below shows two parallel lines, m and n, cut by a transversal, k. Angles A, B, and C are shown in the diagram

Mathematics
1 answer:
puteri [66]3 years ago
8 0

9514 1404 393

Answer:

  B

Step-by-step explanation:

Angles A and C are vertical angles; angles B and C are alternate interior angles. Only line 3 of the proof is in error.

The applicable description is found in choice B.

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Calculate the missing value. (Not drawn to scale.)
True [87]

Answer:5in

Step-by-step explanation:

1. Identify the triangle

you are given a right triangle with the hypotonus missing and are given the side lengths 3 and 4 you know the hypotonus is 5 by the 3,4,5 Pythagorean tripe, if you do not notice this it can be solved with the Pythagorean theorem a^2+b^2=c^2

2. solve (if not done with 3,4,5 triple)

3^2+4^2=c^2

9+16=c^2

25=c^2

5=c | square root both sides to cancel the square

5 0
1 year ago
Write 9.9.9.9<br> 9.9.9 using exponential notation
kramer
9x9x9x9x9x9x9 = 9^7

Hope you’re having a positive day
5 0
3 years ago
9g=56 <br><br> (2 decimal place)
MaRussiya [10]

Answer:

6.2

Step-by-step explanation:

9g=56

9g/9=56/9

g=6.2

5 0
3 years ago
If α, β are the zeroes of the polynomials f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) =
Pavlova-9 [17]

Answer:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

Step-by-step f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)explanation:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

4 0
3 years ago
Can someone help me solve this?
serg [7]

Answer:

(a) y = -3/5 x + 13/5

(b) y = 5/3 x + 1/3

Step-by-step explanation:

(a) The slope of the tangent line is dy/dx.  Use implicit differentiation:

x² + y² + 4x + 6y − 21 = 0

2x + 2y dy/dx + 4 + 6 dy/x = 0

2x + 4 + (2y + 6) dy/dx = 0

x + 2 + (y + 3) dy/dx = 0

(y + 3) dy/dx = -(x + 2)

dy/dx = -(x + 2) / (y + 3)

At the point (1, 2), the slope is:

dy/dx = -(1 + 2) / (2 + 3)

dy/dx = -3/5

Using point-slope form of a line:

y − 2 = -3/5 (x − 1)

Simplifying to slope-intercept form:

y − 2 = -3/5 x + 3/5

y = -3/5 x + 13/5

(b) The normal line is perpendicular to the tangent line, so its slope is 5/3.  It also passes through the point (1, 2), so point-slope form of the line is:

y − 2 = 5/3 (x − 1)

Simplifying to slope-intercept form:

y − 2 = 5/3 x − 5/3

y = 5/3 x + 1/3

4 0
3 years ago
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