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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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Graph the equation y=-x. is (5,5) on the line? what is the slope of the line?

bezimeni [28]

You'll actually need to graph for the first segment
but the second part can be solved with algebra

The equation of the line is y(x)=-x

if we test the co0ordinates we can find if the point lies on the line

(5,5)
y(x)=-x

Using the x co-ordinate
y(5)= -5
y=-5

Leaving us with (5,-5)

Because (5,-5) is not (5,5)
This is not a point on the line

The slope of the line is -1

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3 years ago

0.6 litres in the ratio 7:5

Sedaia [141]

Can you Elaborate your question?

5 0

2 years ago

A runner is running a 10 km race. It takes her 17.5 minutes to reach the 2.5 km mark. At that rate, how long will it take her to

Alinara [238K]

Answer:

At her current rate it will take here 70 minutes to finish the race

3 0

3 years ago

Read 2 more answers

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago

Given the graph of a function f. Identify the function by name. Then Graph, state domain & range in set notation:A) f(x) +2B

WARRIOR [948]

The function in the graph has the name of square function.

The domain of a function is all values of x the function can have. The domain of this function is all real numbers:

$\mleft\lbrace x\in\R\mright\rbrace$

The range of a function is all values of y the function can have. The range of this function is all positive numbers, including zero:

$\mleft\lbrace y\in\R\mright|y\ge0\}$

In order to graph f(x) + 2, we just need to translate the graph 2 units up. To find the new points, we need to increase all y-coordinates by 2:

(-2, 6), (-1, 3), (0, 2), (1, 3), (2, 6)

Domain: {x ∈ ℝ}

Range: {y ∈ ℝ | y ≥ 2}

Then, in order to graph f(x) - 2, we just need to translate the graph 2 units down. To find the new points, we need to decrease all y-coordinates by 2:

(-2, 2), (-1, -1), (0, -2), (1, -1), (2, 2)

Domain: {x ∈ ℝ}

Range: {y ∈ ℝ | y ≥ -2}

4 0

1 year ago

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

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