Answer:
See explanation below
Step-by-step explanation:
Area = Area of triangle + Area of the semi circle
Area of triangle = 1/2bh
Area of triangle = 1/2 *(5/2)(4)
Area of triangle = 1/2(2.5)*4
Area of triangle = 2 * 2.5
Area of triangle = 5cm^2
Area of semicircle = πr²/2
r = 2.5cm
Area of semicircle =(3.14)(2.5)²/2
Area of semicircle = = 3.14 * 4.5
Area of semicircle = 9.8125cm^2
Area pf the figure = 5+9.8125
Area pf the figure = 14.8125cm^2
For the perimeter
Perimeter of the semicircle = 2πr
Perimeter of the semicircle = 2(3.14)(2.5)
Perimeter of the semicircle = 15.7cm
Perimeter of the semicircle = = 15.7 + 5 + 5
Perimeter of the semicircle = 25.7cm
Answer: 4.00
Explanation
To find the maximum value for x, the hypotenuse, with a given of an angle 30° and a height of 2 m, we can see that 30° is directly opposite the side of 2 m, in which case we may use the sin of 30° to find the hypotenuse.
Remember that the sin = opposite/hypotenuse, so sin30° = 2/x
1) sin30 = 2/x
2) 1/2 = 2/x
3) x = 4
Hope this helps :)
Answer:
f(x)=-18x^2
Step-by-step explanation:
Given:
1+Integral(f(t)/t^6, t=a..x)=6x^-3
Let's get rid of integral by differentiating both sides.
Using fundamental of calculus and power rule(integration):
0+f(x)/x^6=-18x^-4
Additive Identity property applied:
f(x)/x^6=-18x^-4
Multiply both sides by x^6:
f(x)=-18x^-4×x^6
Power rule (exponents) applied"
f(x)=-18x^2
Check:
1+Integral(-18t^2/t^6, t=a..x)=6x^-3
1+Integral(-18t^-4, t=a..x)=6x^-3
1+(-18t^-3/-3, t=a..x)=6x^-3
1+(6t^-3, t=a..x)=6x^-3
That looks great since those powers are the same on both side after integration.
Plug in limits:
1+(6x^-3-6a^-3)=6x^-3
We need 1-6a^-3=0 so that the equation holds true for all x.
Subtract 1 on both sides:
-6a^-3=-1
Divide both sides by-6:
a^-3=1/6
Raise both sides to -1/3 power:
a=(1/6)^(-1/3)
Negative exponent just refers to reciprocal of our base:
a=6^(1/3)
You'll have to be more specific, like what is each side, as in height, width or length. Sorry!
