$\frac{cos-sin^{2}+1 }{sin(1+cos)}\\\\=\frac{cos - sin^{2}+sin^{2}+cos^{2} }{sin(1+cos)}\\\\=\frac{cos+cos^{2} }{sin(1+cos)}\\\\=\frac{cos(1+cos)}{sin(1+cos)}\\\\=\frac{cos}{sin}\\\\=cot$

( 1 = sin² + cos²)

Step-by-step explanation:

4 0

3 years ago

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Show that AC and BD bisect each other if A(–3, –5), B(2, –3), C(3, 5), and D(–2, 3). Find AC and BD

Mazyrski [523]

Answer

Multipy all of them

Step-by-step explanation:

mulitiply all of them

7 0

3 years ago

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Now try 50 for x. If x=50,what I'd y when y=x-20? Does that work with the other equation?

SSSSS [86.1K]

X= 50
Replace the x with the given number.
Y= 50-20
Y= 30

5 0

3 years ago

Match each equation to its factorized version and solution.

bagirrra123 [75]

Answer:

$24x -{6x}^{2} = 0 \: matches \: with \: 6x(4 - x) \: and \: the \: solution \: x = 0 \ \: or \: \: x = 4$

$2 {x}^{2} + 6x = 0 \: matches \: with \: 2x(x + 3) = 0 \: and \: the \: solution \: x = 0 \: or \: x = - 3$

$4x - {x}^{2} = 0 \: matches \: with \: x(4 - x) = 0 \: and \: the \: solution \: x = 0 \: or \: x = 4$

$14x - {7x}^{2} = 0 \: matches \: with \: 7x(2 - x) = 0 \: and \: the \: solution \: x = 0 \: or \: x = 2$

5 0

3 years ago