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algol13
3 years ago
6

Two angles that are directly across from each other when two lines intersect. supplementary angles complementary angles adjacent

angels vertical angles
Mathematics
2 answers:
tatuchka [14]3 years ago
8 0

Answer:

Step-by-step explanation:

Two angles that are directly across from each other when two lines intersect are called vertical angles.  Vertical angles are ALWAYS congruent to each other. Very good thing to know when it comes to proving triangle congruency.

Tanzania [10]3 years ago
4 0

Answer:

If two lines intersect each other, then the vertically opposite angles are equal. Theorem 6.1 : If two lines intersect each other, then the vertically opposite angles are equal. In the statement above, it is given that 'two lines intersect each other'. So, let AB and CD be two lines intersecting at O

Step-by-step explanation:

You might be interested in
Please show work for both problems
pentagon [3]

Answer:

1) x = 25    2) x = 23; m∠CBD = 80 degrees

Step-by-step explanation:

1) 5x + 15 = 6x - 10           Set both equations equal to each other

- 5x         - 5x                  Subtract 5x from both sides

15 = x - 10

+ 10    + 10                       Add 10 to both sides

25 = x

2) 2x + 14 + x + 7 = 90         Set the equations equal to 90

   3x + 21 = 90                     Combine like terms

         - 21  - 21                      Subtract 21 on both sides

3x = 69                                 Divide both sides by 3

x = 23

Plug 23 into the equation

2(23) + 14 = 60

5 0
4 years ago
Help me please explain is not needed but would be appreciated
TiliK225 [7]

Answer:

<h2>B = 18°</h2>

Step-by-step explanation:

To find angle B we use tan

tan ∅ = opposite / adjacent

From the question

AC is the opposite

BC is the side adjacent to angle B

So we have

tan B = AC / BC

tan B = 6/9

tan B = 1/3

B = tan-¹ 1/3

B = 18.43°

B = 18° to the nearest hundredth

Hope this helps you

8 0
3 years ago
My age is the difference between twice my age in 4 years and twice my age 4 years ago how old am I?
Llana [10]

Answer: 16

Step-by-step explanation:

let a = your present age

a = 2(a+4) - 2(a-4)

a = 2a + 8 - 2a + 8

a = 2a - 2a + 8 + 8

a = 16 yrs is your age

6 0
3 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0
3 years ago
Are the fractions equivalent?
dangina [55]

Answer:

7. no

8.yes

9.yes

10.no

I hope this helps!

4 0
3 years ago
Read 2 more answers
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