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yulyashka [42]
3 years ago
14

Through which pairs of points is the slope negative? Select all that apply. a. (3, 6) and (10, 6)

Mathematics
1 answer:
xeze [42]3 years ago
5 0
Trial and error on this one try each pair till you find negative slopes
C and D
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Someone please help!!!!!! trigonometry- trig equations!!!
Andru [333]

Answer:

1. 0, 2 π , π , 2 π/3 , 4 π/3

2. 7 π/6 , 11 π/6

Step-by-step explanation:

1. we want to find the value of theta

Let us represent theta by x for ease of documentation

So we have ;

sin 2x + sin x = 0

Mathematically;

sin 2x = 2sinxcosx

So;

2 sin x cos x + sin x = 0

sin x (2 cos x + 1) = 0

sin x = 0

or

2cos x + 1 = 0

x = arc sin (0)

within the given range;

x = theta = 0 , 2 π , π

For;

2 cos x + 1 = 0

2cos x = -1

cos x = -1/2

cos x = -0.5

x = arc cos (-0.5)

x = 2 π/3 , 4 π/3

2. cos 2 x = 1 + sin x

Mathematically;

cos 2 x = 1-2 sin^2x

let sin x = y

1-2y^2 = 1 + y

y = -2y^2

1 = -2y

y = -1/2

y = -0.5

sin x = y

sin x = -0.5

x = arc sin (-0.5)

x = 7 π/6 , 11 π/6

4 0
3 years ago
Tell whether the equation has one solution, infinite many solutions, or no solutions. 5w - 6w = -w + 5
vfiekz [6]

Answer:

No solution

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\heartsuit$ and $\dia
Gnoma [55]

Answer:

The number of ways to select 2 cards from 52 cards without replacement is 1326.

The number of ways to select 2 cards from 52 cards in case the order is important is 2652.

Step-by-step explanation:

Combinations is a mathematical procedure to compute the number of ways in which <em>k</em> items can be selected from <em>n</em> different items without replacement and  irrespective of the order.

{n\choose k}=\frac{n!}{k!(n-k)!}

Permutation is a mathematical procedure to determine the number of arrangements of <em>k</em> items from <em>n</em> different items respective of the order of arrangement.

^{n}P_{k}=\frac{n!}{(n-k)!}

In this case we need to select two different cards from a pack of 52 cards.

  • Two cards are selected without replacement:

Compute the number of ways to select 2 cards from 52 cards without replacement as follows:

{n\choose k}=\frac{n!}{k!(n-k)!}

{52\choose 2}=\frac{52!}{2!(52-2)!}

      =\frac{52\times 51\times 50!}{2!\times50!}\\=1326

Thus, the number of ways to select 2 cards from 52 cards without replacement is 1326.

  • Two cards are selected and the order matters.

Compute the number of ways to select 2 cards from 52 cards in case the order is important as follows:

^{n}P_{k}=\frac{n!}{(n-k)!}

^{52}P_{2}=\frac{52!}{(52-2)!}

       =\frac{52\times 51\times 52!}{50!}

       =52\times 51\\=2652

Thus, the number of ways to select 2 cards from 52 cards in case the order is important is 2652.

6 0
4 years ago
What approach is being used in terms of the outcome of a students grade?
Butoxors [25]
I do
Gfggfsdyyewweewwwedf
3 0
3 years ago
Read 2 more answers
3 6/12 +4 1/2 tell me what it is
m_a_m_a [10]

Answer:

8

Step-by-step explanation:

3 + 4 = 7

6/12 + 1/2 = 1

7+1 = 8

5 0
3 years ago
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