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2 years ago

The point (6, - 5) is reflected across the y axis what is the new point

Mathematics

1 answer:

jonny [76]2 years ago

8 0

Answer:

(-6, -5)

Step-by-step explanation:

Reflection across the y-axis leaves the point on the same horizontal line, but with the sign of its x-coordinate changed.

(x, y) ⇒ (-x, y) . . . . . reflection across the y-axis

(6, -5) ⇒ (-6, -5)

The image point is (-6, -5).

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Please help!!!!!!!!!!!!!!!!!!!!!!!

egoroff_w [7]

Answer:

about -0.09

Step-by-step explanation:

Plug in -0.09 in a calculator into the inverse tan function to get about -0.09 or what is said to be -0.089 radians or -5.14 degrees

Hope i helped, vote me brainly :)

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3 years ago

What is the area of the triangle? PLSSS help me

AveGali [126]

Answer:

The area of the triangle is $A=6 \:units^2$ .

Step-by-step explanation:

The area A of a triangle is given by the formula $A=\frac{1}{2} bh$ where b is the base and h is the height of the triangle.

From the graph, we can see that the base is 3 units and the height is 4 units. Therefore, the area of the triangle is

$A=\frac{1}{2} \cdot3\cdot 4=\frac{12}{2}=6 \:units^2$

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3 years ago

Can someone help me with this math problem 26.04 divided by -6.2

Anton [14]

Answer:

-4.2

Step-by-step explanation:

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3 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)