All categories

3 years ago

WHO EVER ANSWERS FIRST I WILL GIVE THE CROWN-

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

The answer is C because 5 and 7 have the greatest incline inbetween the angles.

You might be interested in

Write the number in standard form.<br> 7 x 107

laila [671]

Answer:

Step-by-step explanation:

Break apart the equation:

(7 x 100) + (7 x 7)

7 x 100 = 700

7 x 7 =49

700 + 49 = 749

Hope this helps

3 0

3 years ago

Helpp plsss!!!! I've to submit ASAP

alexandr1967 [171]

i don’t rly remember surface area i’m sorry i couldn’t help

8 0

3 years ago

Read 2 more answers

reflect (7,4) across the x-axis then reflect the result across the y-axis what are the coordinates of the final point

Andrews [41]

The coordinates of the final point after the two reflections is (-7, -4)

<h3>How to reflect a point across an axis?</h3>

For a point (x, y), a reflection across the x-axis gives the point (x, -y), while for a point (x, y) a reflection across the y-axis gives (-x, y).

In this case, we have the point (7, 4), first we reflect across the x-axis, so we get the point (-7, 4).

Now we reflect across the y-axis, so the sign of the y-component changes, and we get (-7, -4)

So the coordinates of the final point after the two reflections is (-7, -4).

If you want to learn more about reflections, you can read:

brainly.com/question/4289712

8 0

2 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 :)

5 0

3 years ago

I need help please thank you

jonny [76]

The answr is A sorry if wrong

5 0

3 years ago

Read 2 more answers