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2 years ago

How do I graph this equation?

Mathematics

1 answer:

NARA [144]2 years ago

5 0

Answer:

I'm pretty sure you start off with the dot at -3,

then go up 2 times,

and go right 3 times.

then put the dot there.

then you can put your line down.

Step-by-step explanation:

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Andreas93 [3]

$3+3\sqrt[6]{x}~~ = ~~5\implies \stackrel{\textit{1st step,-3 to both sides}}{3\sqrt[6]{x}~~ = ~~2}\implies \sqrt[6]{x}=\cfrac{2}{3} \\\\\\ (\sqrt[6]{x})^6=\left( \cfrac{2}{3} \right)^6\implies x=\cfrac{2^6}{3^6}\implies x=\cfrac{64}{729}$

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2 years ago

The center of a circle is at the origin on a coordinate grid. A line with a positive slope intersects the circle at (0, 7). Whic

denis23 [38]

The true statement is the circle has the radius equal to 7.

<h3>What is Equation of circle?</h3>

The general equation for a circle is (x-h)² + (y-k)² = r², where ( h, k ) is the center and r is the radius.

As, standard form of equation of circle is

(x-h)² + (y-k)² = r²

At origin, equation of circle will be

(0-h)² + (0-k)² = r²

r² = x² + y²

Now, The line intersects the circle at (0,7)

r² = 0² + 7²

r² =49

r=7.

Hence, the circle has the radius equal to 7.

Learn more about this concept here:

brainly.com/question/11702700

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3 0

1 year ago

A street in San Francisco rises 20 feet for every 100 feet horizontally. What is the measure of the angle of elevation,

SIZIF [17.4K]

Answer:

11°

Step-by-step explanation:

tan ∅ = 20/100 = 0.2

∅ = 11.31°

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2 years ago

The bananas are $.15 per pound and I want to buy 4 pounds, how much would I pay

Paladinen [302]

$.75 bc 15 x 4 is 75 so that would make 75¢ or $.75

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1 year ago

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

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2 years ago