MATHEMATICS HIGH SCHOOL

Solve: sec b cosec b=2 cosec b

Answers

Answer 1
Answer:

Answer:

b = 60° or 300°

Step-by-step explanation:

We can solve the equation sec b csc b=2 csc b in terms of sin and cos as follows:

We can write sec b and csc b in terms of sin and cos:

\sf sec \:b = (1)/(\:cos\: b)

\sf csc b = (1)/(sin \:b)

Substituting these expressions into the equation, we get:

\sf (1)/(cos b) \cdot (1)/(sin b) = 2 \cdot (1)/(sin b)

Multiplying both sides of the equation by sin b cos b, we get:

\sf 1 = 2\cdot (cosb\cdot sinb)/(sinb)

\boxed{\sf (sin b)/(sinb)=1}

so,

\sf 1 = 2\: cos b

Dividing both sides of the equation by 2, we get:

\sf (1)/(2) =cos b

\sf cos 60^\circ = cos b

Since cos is positive in 1st and 4th quadrant, so

\sf cos 60^\circ \:\:or\:\:cos(360^\circ-60^\circ)= cos b

Therefore, the solutions to the equation sec b \csc b=2 \csc b for b is 60° or 300°

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Factor the expression.15n−18

Enter your answers in the boxes to complete the factored expression.

Answers

15n - 18 = 3(5n -6)
I suppose that's all...

Answer:15n - 18 = 3(5n -6)

Step-by-step explanation:

What is the slope of the line that passes through 2,-1 and 2,-5

Answers

Answer:

Undefined.

Step-by-step explanation:

When using the formula y2-y1/x2-x1, the result comes out as -4/0. Any number over a 0 is undefined.

5 sevenths times 3 fourths

Answers

5/7 x 3/4 = (5 x 3)/(7 x 4) = 15/21 = 5/7
the answer would be .5357142857, but converted to a fraction it is 15/28.

If (-3, y) lies on the graph of y = 3-x, then y =

-27
1/27
27

Answers

sub
(x,y)
(-3,y)
x=-3
y=y
y=3-(-3)
y=3+3
y=6

y=6 which is mysteriously not listed

you're going to want to pick

B. 1/27

because

y=3^-3

y=1/(3^3)

y=1/(3 x 3 x 3)

y=1 / 27

i have no clue how this guy got Verrified answer

its a 3 choice question on odyssey ware.

we didn't make it man

Factor completely. n ^4 - 1

Answers

Answer: Hello there!

here we have the equation n^4 - 1

using the relation: (a^(2) - b^(2)) = (a+b)*(a-b)

we can write our equation as:

(n^(4) - 1) = ((n^(2) )^(2) -1^(2) ) = (n^(2) + 1)(n^(2) - 1) = (n^(2) + 1)(n + 1)(n-1)

and (n^2 + 1) has only complex roots, i and - i, then we can factorize this as (n -i)(n + i) = n*n + ni - ni (+i)*(-i) = (n^2 + 1)

then our equation is: (n^(2) + 1)(n + 1)(n-1) =  (n + i)(n - i)(n + 1)(n-1)
(n^4 - 1) = (n^2 - 1) (n^2 +1) = (n - 1) (n + 1) (n^2 + 1)
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