Since she already spent $8 for 2 pounds of the ground coffee, she now has $12 left so then she can only buy only 1 pound of gourmet coffee since she will only have $4 left after buying the first pound.
About 4.1 seconds. How long was the ball in the air? We are told that t represents time in seconds since the ball was thrown, so it started to be 'in the air' at t = 0 To answer the question, then, we need to know the time when it stopped being in the air. We are told that the ball hit the ground. So that's what happened when it stopped being airborne. We need to relate that event to the mathematics we're working with. What can we say about h , the height of the ball when the ball hits the ground? Answer: The height will be 0 when the ball stops being in the air. Now translate this back to the mathematics: The ball is in the air from t = 0 until the time t when h = 0 . Find the time t that makes h = 0 . That means: solve: − 5 t 2 + 20 t + 2 = 0 We can solve this by solving: 5 t 2 − 20 t − 2 = 0 (Either multiply both sides of the equation by − 1 , or add 5 x 2 − 20 x and − 2 to both sides and then re-write it the other way around) That's a quadratic equation, so try to factor first. But don't spend too much time trying to factor, because not every quadratic is easily factorable and that's OK, because we still have the quadratic formula if we need it. We do need it. t = − ( − 20 ) ± √ ( − 20 ) 2 − 4 ( 5 ) ( − 2 ) 2 ( 5 ) = 20 ± √ 440 10 = 20 ± √ 4 ( 110 ) 10 = 20 ± 2 √ 110 10 = 2 ( 10 ± √ 110 ) 2 ( 5 ) = 10 ± √ 110 5 We can see that 10 < √ 110 < 11 . In fact ( 10 + 1 2 ) 2 = 10 2 + 10 + 1 4 = 110.25 Using 10.25 as an approximation for √ 110 , we get : for the solution t = 10 − √ 110 5 we'll get a negative t . That doesn't make sense. The other solution gives t ≈ 10 + 10.25 5 = 20.5 5 = 4.1 seconds. So the ball was in the air from t = 0 until about t = 4.1 . The elapsed time is the difference, 4.1 seconds.
Y = -1 is a horizontal line going through "-1" on the y axis
Note that the point (1,2) is exactly 3 units of distance above the line y = -1
When we reflect across this line, the point (1,2) will just move straight down to exactly 3 units of space below the line y = -1. Since we are not shifting left or right, the x coordinate of our original point will not change. The y coordinate of our original point will now need to be reduced by 6(3 units down to get to the line of reflection and then 3 more down to get to the image location)
The coordinates of the image point will be (1, -4)
Now we need to do the same process with (1, -4) being reflected across y=1
Note (1,-4) is 5 units of distance below the line y = 1 , so we need to reflect the point upward so that the image point is located exactly 5 units of distance above the line y = 1 Again, the x coordinate does not change, and our final image coordinates are (1, 6)
I guess more simply stated, if you're just looking for the number in the green box it would be " 1 " .. Reflecting points across horizontal lines only result in changes of the "y" coordinate since there is no shift left or right.
Answer:
26
Step-by-step explanation:
40-32÷8+5×-2
40-4+5×-2
40-4+-10
36+-10
26
Answer:
WINKEY + D. ...
WINKEY + SPACE. ...
SHIFT + Mouse Click on a taskbar button. ...
CTRL + SHIFT + Mouse Click on a taskbar button. ...
SHIFT + Right Mouse Click on a taskbar button. ...
SHIFT + Right Mouse Click on a grouped taskbar button. ...
CTRL + Mouse Click on a grouped taskbar button. ...
WINKEY + T.
ctrl + X = cut
ctrl + c = copy
make me brainliest
