Here is your answer

REASON:

: The opposite sides of a parallelogram are equal.
So, in above given figure

(measures of opposite sides)




HOPE IT IS USEFUL
The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)
<em><u>Solution:</u></em>
Given that,

<em><u>We have to substitute eqn 1 in eqn 2</u></em>






Substitute x = 2.1925 in eqn 1
y = 2.1925 + 3
y = 5.1925
Substitute x = -3.1925 in eqn 1
y = -3.1925 + 3
y = -0.1925
Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)
The solution would be like this for this specific problem:
√512m³ = √256m² × √2m
= 16 × √2m
I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.
Answer:
64
Step-by-step explanation:
Evaluate the parenthesis before doing the addition, that is
(6 × 9) + (7 + 3)
= 54 + 10
= 64
The answer is
round cake - 82.42 in²
rectangular cake - 114 in²
Round cake:
d = 7 in
r = d/2 = 7 in / 2 = 3.5 in
h = 2 in
The surface are of a cylinder is:
A = 2πr² + 2πrh
The surface are of the round cake (which is actually a cylindrical cake) excluding the bottom is:
A = 2πr² + 2πrh - πr²
A = πr² + 2πrh
A = 3.14 * 3.5² + 2 * 3.14 * 3.5 * 2
= 38.46 + 43.96
= 82.42 in²
Rectangular cake:
w = 6 in
l = 9 in
h = 2 in
The surface are of a rectangle is:
A = 2wl + 2wh + 2lh
The surface are of the rectangular cake excluding the bottom is:
A = 2wl + 2wh + 2lh - wl
A = wl + 2wh + 2lh
A = 6 * 9 + 2 * 6 * 2 + 2 * 9 * 2
= 54 + 24 + 36
= 114 in²
Step-by-step explanation: This answer is not mine but JcAlmighty’s so all credits go to them.