Answer:
56√2 ≈ 79.20
Step-by-step explanation:
The hypotenuse of an isosceles right triangle is √2 times the side length, so the side length here is ...
s·√2 = 28
s = 28/√2
The perimeter is 4 times the side length, so is ...
P = 4s = 4·28/√2
= 4·28·(√2)/2 . . . . . multiply by (√2)/(√2) to rationalize the denominator
= 56√2 ≈ 79.195959
The perimeter of the square is 56√2, about 79.20 units.
1. The shape of cross-section is a circle.
2. The face parallel to ABCD is EFGH. Since this is a a rectangular shape,
A = L*H = 12*6 = 72 cm^2
3. The cross-section parallel to ABC is DEF with h = 12 ft, b= 5ft (where h is the height and b is the base of a right angled triangle).
Area, A = 1/2 *b*h = 1/2*5*12 =30 ft^2
4. Plane BDHF is a rectangle shape whose length is the diagonal of ABCD.
Diagonal BD = sqrt (AB^2+BD^2) = sqrt (8^2+7^2) = 10.63 cm.
Perimeter, P = 2(BD+DH) = 2(10.63+6) = 33.26 cm
Y = -10 and x = -13 (substitute 2y + 7 for x)
Answer:
5
Explanation:
Let the number equal x. Half the number is then
x
2
and the reciprocal of that is
2
x
The reciprocal of the number is
1
x
and half that is
1
2
x
then
2
x
+
1
2
x
=
1
2
4
x
+
x
2
x
2
=
1
2
10
x
=
2
x
2
2
x
2
−
10
x
=
0
2
x
(
x
−
5
)
=
0
Zero is not viable solution as its reciprocal is infinity. The answer is therefore
x
=
5
The answer is false. Good luck with other questions!
