All categories

2 years ago

Using the picture below, which angle is alternate exterior to angle 2?

Mathematics

1 answer:

kenny6666 [7]2 years ago

4 0

Answer:

alternate exterior means outside of the two parallel lines so your answer is 7 because alternate means other side of transversal

You might be interested in

I REALLY NEED HELP ON 40 and 41, 50 POINTS

photoshop1234 [79]

Question 40 is D because from looking at the graph, its is symmetrical and the mean is the average, which usually the middle of the graph, so is the median. The median and mean would be within the 2.1-3 category because its the middle of the graph.

Question 41 is B because every pound is 16 ounces. If you multiply 16 and 12.5, you'll get 200.

3 0

3 years ago

How old am i if 200 reduced by 2 times my age is 16

Rashid [163]

$x-\ you're\ age\\\\
200-2x=16\ \ \ | subtract\ 200\\\\
-2x=-184\ \ \ | divide\ by\ -2\\\\
x=92\\\\
You\ are\ 92\ years\ old.$

3 0

3 years ago

Write a variable expression for this amount. 8 more than twice a number d

jarptica [38.1K]

2n+8 maybe try that i dont get the question..kinda confusing

6 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$