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3 years ago

PLEASE HELP!!! WILL GRANT BRAINLIEST IF CORRECT, NEED NOW

Mathematics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

For each equation, set x to 0. A is correct because if x=o, then y will equal 38.910.

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In the plane figure, only downward motion (movement leaving you relatively lower than where you were) is allowed. Find the total

bixtya [17]

Plane figures are shapes with straight sides. Considering the movement to be lower than the initial path, there are eleven (11) paths from A to B.

A plane figure is given shape with straight sides. And these sides connect each edge of the figure one to another. Examples are triangles, quadrilaterals, polygons, etc.

A path is a connection or link between two reference points. This could be a footpath or a motorable path or other forms of the path.

Considering the given plane figure in the question, it can be observed that the total number of paths from A to B with respect to the given condition is eleven.

For more clarifications on paths around a given plane figure, visit: brainly.com/question/4871149

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8 0

2 years ago

A line has a slope of 2 and a y intercept of (0,-4) what is the value of y when x=6

Pavel [41]

First we set up an equation with the givens (slope and y intercept):
y = 2x -4
next plug in 6 for x.
y = (2x6) - 4
y = 12 - 4

And this will give us the value of y: 8.

3 0

3 years ago

V = the square root of 2as Solve for s

Gnom [1K]

As, v=√2as
Squaring both sides,
v²=2as
∴ s= v²/2a

6 0

3 years ago

Please help with this

notsponge [240]

Answer:

Step-by-step explanation:

By the Pythagorean theorem:

$18^2+24^2=x^2$

Therefore, answer choice D is correct. Hope this helps!

8 0

4 years ago

Given the function f(x) x^2-2x/x^3 +9x^2-10x Find any holes, vertical asymptotes, and horizontal asymptotes there may be.

yulyashka [42]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ :

Holes: Since $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ reduces to $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ , then there is a hole at $x=0$ as $x$ exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).

Vertical Asymptotes: If we further reduce $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ to $f(x)=\frac{x-2}{(x+10)(x-1)}$ , then we see that there are vertical asymptotes at $x=-10$ and $x=1$

Horizontal Asymptotes: As the degree of the numerator is less than the degree of the numerator ( $2$ ), then there is a horizontal asymptote at $y=0$

4 0

2 years ago