Answer:
![\huge\boxed{Q = 110 \textdegree, \ ST = 9 \ \text{ft}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7BQ%20%3D%20110%20%5Ctextdegree%2C%20%5C%20ST%20%3D%209%20%5C%20%5Ctext%7Bft%7D%7D)
Step-by-step explanation:
When we reflect and transform a figure, the angle lengths nor the side lengths are touched. The side length only changes if we dilate or stretch the figure, and the angle length only changes if we stretch the figure.
Therefore, we know the information is going to be the exact same. We just have to figure out what corresponds to what from JKLM to QTSR.
If we reflect a figure across a horizontal reflection line through its center, the figure will just flip sides. M will be J, L will be K, etc.
When we translate this down, nothing changes except its position. So we can pretend these two shapes are right on top of each other for now.
When we move these right on top of each other, we can see that Angle J overlaps with Angle Q. Since they don't have any weird intersects, we know that angle J will be equal to Angle Q. Since we already know J is 110°, we know Q is also 110°.
When we move it on top, we also see that KL overlaps with ST perfectly. Since we know KL is 9, that must mean ST is also 9.
Hope this helped!
Answer:
i need a pic
:)
Step-by-step explanation:
2
![2x^2+7x-10 = x+5 2x^2+7x-15 y2x^2+6x-15](https://tex.z-dn.net/?f=2x%5E2%2B7x-10%20%3D%20x%2B5%0A%0A2x%5E2%2B7x-15%0A%0Ay2x%5E2%2B6x-15%20)
Suggesting that you want this in standard form, in terms of quadratic equations, you would technically follow a process similar if not almost exactly like the 2 - step equation method with the exception of separating the (x)s and the equations to find x and then plug it in and what-not.
With that being said you would subtract 5 in (x+5) from said 5 in the second equation and -10 in the first equation in order to get 2x^2+7x-15, you would continue to do the same for the x by subtracting it from both ends making the 7x a 6x because there is a 1 at the beginning of each x if there is no number that is shown already. Which finally gives you the equation (y= 2x^2+6x-15)
Bernardo travels the same distance at 25mph as he does at 50mph. However, since 25mph is only half of 50 mph, he must travel twice as long at 25mph. If you call the time he traveled 50mph "t", then
<span>t+2t=3 </span>
<span>3t=3 </span>
<span>t=1 </span>
<span>This means he traveled 1 hour at 50mph. In this time, he traveled 50 miles. He traveled the same distance at 25mph, so his total distance was </span>
<span>50miles+50miles=100miles </span>
<span>so the round trip was 100 miles.</span>