Answer: 1. True 2. True 3. False. Number 4 True
Step-by-step explanation: PLS GIVE ME BRAINIEST
Answer: 52 degrees
Step-by-step explanation:
most likley like if correct
Answer:
Step-by-step explanation:
So, we have-
(6 x .1) + (3 x .01) + (2 x .001)
So, we will solve inside of the parenthesis first.
6 x .1 = <u>.6</u>
So, we got that settled next-
3 x .01 = <u>.03</u>
We got that settled last-
2 x .001 = <u>.002</u>
Lastly, we add them all.
.6 + .03 + .002 = .632
Now we compare.
.632 > .629
So, there is your answer!
Hope this helps!
The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?
Given : - Square ABCD with side 3. E and F as midpoints.
To find : - area of EBFD
Solution : - We have, area of square ABCD = 3 x 3 = 9 units.
Thus, (ar)EBFD = ar ABCD - ar DAE - arDCF
arDAE = 1/2 x base x height
=1/2 x 1.5 x 3 ( AE is 1/2 of AB = 1.5, DA is altitude)
= 2.25
arDFC = 1/2 x base x height
= 1/2 x 1.5 x 3 (FC is 1/2 of BC, DC is altitude)
= 2.25
Thus, (ar) EBFD = arABCD - arDAE - arDCF
= 9 - 2.25 - 2.25
= 4.5 units.
Thus, area of quad EBFD is 4.5 units.