Hello! I can help you with this!
A. Okay. So, to find the area of the square all you have to do is the number square, or multiply the number by itself, in this case. 3.5. 3.5^2 is 12.25. The area of the square is 12.25 square meters large.
B. To find the area of the circle, you do pi*r^2. Radius is half the the diameter. Half of 4 is 2. 2^2 is 4. To multiply by pi, we can just do it as 3.14. 4 * 3.14 is 12.56. The circle is 12.56 square meters large.
C. The circle garden would give her the most space to plant, because 12.56 is larger than 12.25 by 0.31. The circle garden 0.31 square meters larger than the square garden.
Answer:
1
+
sec
2
(
x
)
sin
2
(
x
)
=
sec
2
(
x
)
Start on the left side.
1
+
sec
2
(
x
)
sin
2
(
x
)
Convert to sines and cosines.
Tap for more steps...
1
+
1
cos
2
(
x
)
sin
2
(
x
)
Write
sin
2
(
x
)
as a fraction with denominator
1
.
1
+
1
cos
2
(
x
)
⋅
sin
2
(
x
)
1
Combine.
1
+
1
sin
2
(
x
)
cos
2
(
x
)
⋅
1
Multiply
sin
(
x
)
2
by
1
.
1
+
sin
2
(
x
)
cos
2
(
x
)
⋅
1
Multiply
cos
(
x
)
2
by
1
.
1
+
sin
2
(
x
)
cos
2
(
x
)
Apply Pythagorean identity in reverse.
1
+
1
−
cos
2
(
x
)
cos
2
(
x
)
Simplify.
Tap for more steps...
1
cos
2
(
x
)
Now consider the right side of the equation.
sec
2
(
x
)
Convert to sines and cosines.
Tap for more steps...
1
2
cos
2
(
x
)
One to any power is one.
1
cos
2
(
x
)
Because the two sides have been shown to be equivalent, the equation is an identity.
1
+
sec
2
(
x
)
sin
2
(
x
)
=
sec
2
(
x
)
is an identity
Step-by-step explanation:
The answer is 3.814 but if you wanna round to the nearest hundredth then it’s 3.81
Answer: 75+30 = 15 x 7
Step-by-step explanation:
The given expression is 75+30 (=105) which defines the sum of 75 and 30.
Prime factorization of 75 and 30 are as below:
75 = 5 x 5 x 3
30 = 5 x 3 x 2
GCD (75,30) = 5x 3 = 15 [Note: GCD = Greatest common divisor]
Consider 75+30 = (15 x 5) + (15 x 2) [75 = 15 x 5 and 30= 15 x 2]
= 15 (5+2) [taking 15 as common ]
= 15 x (7)
(=105)
So, 75+30 which is sum of the numbers and it is expressed as 15 x 7 which a product of their GCF.