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mihalych1998 [28]
3 years ago
6

In the picture below, line PQ is parallel to line RS, and the lines are cut by a transversal, line TU. The transversal is not pe

rpendicular to the parallel lines. Note: Figure is not drawn to scale. Which of the following are congruent angles?
Mathematics
1 answer:
natta225 [31]3 years ago
4 0

Answer:<W=<E the answer is W and E

Step-by-step explanation:

i got it right on study island

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I'm having a bit of trouble finding out how to find the Domain of this function. Help please!
BabaBlast [244]

In general, the domain is the set of all x-values for the graph.  

The issue here is that this isn't the graph of a function.  A function has at most one y-value for each x-value and this graph has an infinite number of y-values for the single x-value of 1.

So, either your teacher is wanting you to say the domain is {1}, because that's the only x-value used by the graph, or they're wanting you to say this is a trick question, because this isn't the graph of a function.

The range is the set of all y-values, which is -9<y<9, but again, do they intend this to be a trick question?

6 0
3 years ago
The lengths of a triangle are 12 and 23 meters find the range of possible lengths for the third side
sergij07 [2.7K]

Answer: 11 < x < 35

suppose: the length of the third side is x

because x is the third side of a triangle

=> 23 - 12 < x < 23 + 12

⇔ 11 < x < 35

Step-by-step explanation:

6 0
3 years ago
Find the area of the shaded sector. Leave your answer in terms of π.
bearhunter [10]

Answer:

d=10m

r=10/2=5m

The Area is:

(πr^{2}) /2

(π5^2) / 2

=25π/2

=12.5π

8 0
3 years ago
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6(5x-3) what is the answer to this
Zarrin [17]

Answer:

= 30x − 18

Step-by-step explanation:

3 0
3 years ago
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Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
3 years ago
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