All categories

2 years ago

Do alternate interior angles add to 180 degrees

Mathematics

2 answers:

elena-14-01-66 [18.8K]2 years ago

6 0

No alternate interior angles are equal.

Anna007 [38]2 years ago

3 0

Yes, they can add up to 180 degrees. Say one angle is 65 degrees, and the other one is 115 degrees. That would equal to 180 degrees.
Hope this helps.

You might be interested in

Find the acute angles between the curves at their points of intersection. (the angle between two curves is the angle between the

amm1812

At point of intersection the two equations are equal,
hence, 6x³ =6x²
6x³-6x²=0
6x²(x-1)=0 , the values of x are 0 and 1
The points of intersection are therefore, (0,0) and (1,6)
To find the slopes of the tangents at the points of intersection we find dy/dx
for curve 1, dy/dx=12x, and the other curve dy/dx=18x²
At x=0, dy/dx=12x =0, dy/dx=18x² = 0, hence the angle between the tangents is 0, because the tangents to the two curves have the same slope which is 0 and pass the same point (0,0) origin.
At x=1, dy/dx =12x = 12, dy/dx= 18x² =18, Hence the angle between the two tangents will be given by arctan 18 -arctan 12
= 86.8202 - 85.2364 ≈ 1.5838, because the slope of the lines is equal to tan α where α is the angle of inclination of the line.

5 0

2 years ago

The question is on the picture

Aleonysh [2.5K]

Answer:

The third one is the correct answer!

Because:

there are 2 plots on 39
there is 1 plot on 42
1 plot on 44
1 plot on 45
1 plot on 46
3 plots on 48
and 1 plot on 51

No more no less.

Hope this helped!

6 0

2 years ago

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.