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2 years ago

Show whether or not y=x+3 is tangential to the curve y^2=x

1 answer:

3 0

The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.

y² = x

Take the derivative of both sides with respect to x, assuming y = y(x) :

2y dy/dx = 1

dy/dx = 1/(2y)

Solve for y when dy/dx = 1 :

1 = 1/(2y)

2y = 1

y = 1/2

When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.

This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :

x = y² ⇒ y = y² + 3 ⇒ y² - y + 3 = 0

has no real solution for y.

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What number has the same value as the expression |-12|

maksim [4K]

Answer: 12

Step-by-step explanation:

|-12| |?| the lines on the outside mean that the answer has to be the absolute value examples:

|-3|=3

|2|=2

|-72|=72

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3 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST !!

kati45 [8]

Answer:

angle HIK and angle GFD

Step-by-step explanation:

pretty sure thats right sry if its wrong have a nice day :)

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2 years ago

What is the exact value of sin(theta + beta)?

NISA [10]

Answer: $\frac{-3\sqrt{13}+4\sqrt{3}}{20}$

This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.

sqrt = square root

=======================================================

Explanation:

Angle theta is between pi and 3pi/2. This places the angle in quadrant Q3 where both cosine and sine are negative

Use the pythagorean trig identity to get the following:

$\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1\\\\\sin^2 \theta + \frac{3}{16} = 1\\\\\sin^2 \theta = 1 - \frac{3}{16}\\\\\sin^2 \theta = \frac{16}{16} - \frac{3}{16}\\\\\sin^2 \theta = \frac{16-3}{16}\\\\\sin^2 \theta = \frac{13}{16}\\\\\sin \theta = -\sqrt{\frac{13}{16}} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -\frac{\sqrt{13}}{\sqrt{16}}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\$

Angle beta is in Q1 where sine and cosine are positive.

Draw a right triangle with legs 3 and 4. The hypotenuse is 5 through the pythagorean theorem. In other words, we have a 3-4-5 right triangle.

Since $\tan \beta = \frac{3}{4}$ , this means $\sin \beta = \frac{3}{5} \ \text{ and } \ \cos \beta = \frac{4}{5}$

Use these ideas:

sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent

In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.

-------------------------------------

To recap:

$\cos \theta = -\frac{\sqrt{3}}{4}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\\cos \beta = \frac{3}{5}\\\\\sin \beta = \frac{4}{5}\\\\$

They lead to this

$\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -\frac{\sqrt{13}}{4} * \frac{3}{5} - \left(-\frac{\sqrt{3}}{4}\right) * \frac{4}{5}\\\\\sin\left(\theta + \beta\right) = -\frac{3\sqrt{13}}{20}+\frac{4\sqrt{3}}{20}\\\\\sin\left(\theta + \beta\right) = \frac{-3\sqrt{13}+4\sqrt{3}}{20}\\\\$