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3 years ago

How is naming a line segment different from naming a line?

1 answer:

Studentka2010 [4]3 years ago

7 0

A line is infinite, it has no beginning or end. it's length is infinite.
It can only be named by the equation or a variation of y=mx+c where m is the gradient and c is the y intercept

A line segment has a specific start and end, and a length.
It can be expressed giving two sets of coordinates, (the start and finish)

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I need help if you get it right I will mark you

pav-90 [236]

Answer:

%30 off is the better price

Step-by-step explanation:

firstly we know that 2 - 12 is equal to 10 so that is $10.00 there from that coupon

to find out the second value, we first devide by the total amount (%100 in this case)

12 / 100 = 10.12

then we multiply by the partial amount (%30)

0.12 * 30 = 3.6

convert that into a dollar amount and we get the total discount in dollars

3.60 = $3.60

AND

$3.60 > $2.00

hope this helps ya, take care mate :D

6 0

3 years ago

Read 2 more answers

There are two coins in a bag. for coin 1, p(h) = ½ and for coin 2, p(h) = 1/3. your friend chooses one of the coins at random an

lana [24]

The answer is 3 to this question

8 0

3 years ago

If 2tanA=3tanB then prove that, tan(A+B)= 5sin2B/5cos2B-1

Fed [463]

By definition of tangent,

tan(A + B) = sin(A + B) / cos(A + B)

Using the angle sum identities for sine and cosine,

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

yields

tan(A + B) = (sin(A) cos(B) + cos(A) sin(B)) / (cos(A) cos(B) - sin(A) sin(B))

Multiplying the right side by 1/(cos(A) cos(B)) uniformly gives

tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) tan(B))

Since 2 tan(A) = 3 tan(B), it follows that

tan(A + B) = (3/2 tan(B) + tan(B)) / (1 - 3/2 tan²(B))

… = 5 tan(B) / (2 - 3 tan²(B))

Putting everything back in terms of sin and cos gives

tan(A + B) = (5 sin(B)/cos(B)) / (2 - 3 sin²(B)/cos²(B))

Multiplying uniformly by cos²(B) gives

tan(A + B) = 5 sin(B) cos(B) / (2 cos²(B) - 3 sin²(B))

Recall the double angle identities for sin and cos:

sin(2x) = 2 sin(x) cos(x)

cos(2x) = cos²(x) - sin²(x)

and multiplying uniformly by 2, we find that

tan(A + B) = 10 sin(B) cos(B) / (4 cos²(B) - 6 sin²(B))

… = 10 sin(B) cos(B) / (4 (cos²(B) - sin²(B)) - 2 sin²(B))

… = 5 sin(2B) / (4 cos(2B) - 2 sin²(B))

The Pythagorean identity,

cos²(x) + sin²(x) = 1

lets us rewrite the double angle identity for cos as

cos(2x) = 1 - 2 sin²(x)

so it follows that

tan(A + B) = 5 sin(2B) / (4 cos(2B) + 1 - 2 sin²(B) - 1)

… = 5 sin(2B) / (4 cos(2B) + cos(2B) - 1)

… = 5 sin(2B) / (4 cos(2B) - 1)

as required.

5 0

2 years ago

A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.

kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is y=3x−13

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is 13 , the slope of a line perpendicular to line BC is −3 . The equation of altitude 3 is y=−3x−1

The orthocenter is the point where all three altitudes intersect.

x=2

y=3x−13

y=−3x−1

Use substitution to solve the first two equations y=3(2)−13=−7

The orthocenter is the point (2,−7)

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1

−7=−6−1

−7=−7 it works.

3 0

3 years ago

Find the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2). Can you find another answer?

Margarita [4]

It is given that two vertices of square are (0,0) and (4,2).

Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.

1. By Supposing that these two are adjacent vertices of Square

The third vertex will be at (-4,2) which lies in third quadrant.

Suppose the coordinate of fourth vertex be (x,y).

Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).

Mid point of line joining (x,y) and (0,0) is (x/2,y/2).

Since diagonals of square bisect each other,

∵ x/2=0

⇒x=0

and

y/2=2

⇒y=4

So, The Coordinate of fourth vertex is (0,4).

Now coming back to second condition if these are two opposite vertex of Square.

Let the third coordinate be (a,b).

Length of diagonal= $\sqrt{(4-0)^2+(2-0)^2}=\sqrt20=2\sqrt5$

Now,let side of Square be A.

Then length of Diagonal of square =√2 A

⇒√2 A=2√5

⇒A =√10

As third vertex is (a,b).

Using distance formula

a² + b²=10 -------------(1)

(a-4)²+ (b-2)²=10 --------------(2)

Solving expression (1) and (2), we get

⇒a²+ b²=(a-4)² +(b-2)²

⇒2a + b =5

⇒b=5-2a

Putting the value of b in (1),we get

⇒a² +(5-2a)²=10

⇒a²+25+4a²-20a =10

⇒5a²-20a+15=0

⇒a² - 4a + 3=0

Splitting the middle term,we get

⇒(a-3)(a-1)=0

⇒a=3 ∧ a=1

we get b=5-2×1=3 and b=5-2×3=5-6=-1

So,the other vertex are (1,3) and(3,-1).

3 0

3 years ago