Simplify 3(2x-2)-2(x-2) A.3x-3 B.3x-4 C.4x-2 D.4x-4

Mathematics

2 answers:

Black_prince [1.1K]3 years ago

5 0

The answer is c 4x-2

Artemon [7]3 years ago

4 0

C 4x-2 that’s the answer

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9. Given the point (6,-8) values of the six trig function.

lozanna [386]

Answer:

Here's what I get.

Step-by-step explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10

$\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}$

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

$\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}$

3 0

2 years ago

Which of the following is equal to the expression below?

GuDViN [60]

256/3.hope this helps

6 0

2 years ago

Find the unknown side length x.

Ganezh [65]

Answer:

x = $\sqrt{65}$

Step-by-step explanation:

The altitude of the triangle must equal 4 because it is a right triangle with another leg of 3 and a hypotenuse of 5 (reason is due to the pythagorean theorem)

Therefore, the triangle on the right has legs of 4 and 7 and, again, use the phythagorean theorem to find 'x' (the hypotenuse)

$4^{2}$ + $7^{2}$ = $x^{2}$