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Black_prince [1.1K]
3 years ago
9

Find the equation of the line passing through the points A (2,1) B (7,9)

Mathematics
1 answer:
Bezzdna [24]3 years ago
7 0
m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{9 - 1}{7 - 2} = \frac{8}{5} = 1\frac{3}{5}
              y - y₁ = m(x - x₁)
               y - 1 = 1³/₅(x - 2) Point - Slope Form
               y - 1 = 1³/₅(x) - 1³/₅(2)
               y - 1 = 1³/₅x - 3¹/₅
                 + 1           + 1
                    y = 1³/₅x - 2¹/₅ Slope - Intercept Form
         -1³/₅x - y = 1³/₅x - 1³/₅x - 2¹/₅
          -1³/₅x - y = -2¹/₅
     -1(-1³/₅x - y) = -1(-2¹/₅)
-1(-1³/₅x) + 1(y) = 2¹/₅
           1³/₅x - y = 2¹/₅ Standard Form
         1³/₅(0) - y = 2¹/₅
                0 - y = 2¹/₅
                    -y = 2¹/₅
                    -1    -1
                     y = -2¹/₅ Y - Intercept
               (x, y) = (0, -2¹/₅)
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Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

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2 years ago
Solve the system of two linear inequalities graphically.
Sladkaya [172]

Answer:

See attachment

Step-by-step explanation:

Isolate y in the first inequality:

2x + 4y < -16\\4y < - 16 -2 x\\y < -4 -\frac{1}{2} x\\y < -\frac{x+8}{2} \\

Now, with both x and y inequalities found, graph it.

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2 years ago
Please help me out :P
Veseljchak [2.6K]

cos θ = \frac{-4\sqrt{65} }{65}, sin θ = \frac{-7\sqrt{65} }{65}, cot  θ  = 4/7, sec  θ = \frac{-\sqrt{65} }{4}, cosec  θ  = \frac{-\sqrt{65} }{7}

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = \sqrt{7^{2} + 4^{2} } = \sqrt{49+16} = \sqrt{65}

sin θ = opposite/ hyp = \frac{7}{\sqrt{65} }

Rationalize, \frac{7}{\sqrt{65} } x \frac{\sqrt{65} }{\sqrt{65} } = \frac{7\sqrt{65} }{65}

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - \frac{7\sqrt{65} }{65}

cos   θ  = adj/hyp = \frac{4}{\sqrt{65} }

By rationalizing and knowing that cos  θ  is negative, cos θ  = -\frac{-4\sqrt{65} }{65}

cot θ  = 1/tan θ  = 1/7/4 = 4/7

sec θ  = 1/cos θ  = 1/\frac{4}{\sqrt{65} } = -\frac{-\sqrt{65} }{4}

cosec θ  = 1/sin θ  = 1/\frac{\sqrt{65} }{7} = \frac{-\sqrt{65} }{7}

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

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Answer:

the answer . is 2y

Step-by-step explanation:

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