Answer:
51.9 cm²
Step-by-step explanation:
From the diagram attached,
Area of the white region(segment)(A) = Area minor sector- area of the triangle
A = (πr²∅/360°)-(1/2r²sin∅)............... Equation 1
Where r = radius of the circle, Ф = reflex angle formed at the center of the circle, π = pie
From the question,
Given: r = 7 cm, Ф = 150°
Constant: π = 22/7
Substitute these values into equation 1
A = [(22/7)×7²×150/360]-[(1/2)×7²×sin150]
A = 64.17-12.25
A = 51.92
A = 51.9 cm²
First, you want to establish your equations.
L=7W-2
P=60
This is what we already know. To find the width, we have to plug in what we know into P=2(L+W), our equation to find perimeter.
60=2(7W-2+W)
Now that we only have 1 variable, we can solve.
First, distribute the 2.
60=14W-4+2W
Next, combine like terms.
60=16W-4
Then, add four to both sides.
64=16W
Lastly, divide both sides by 16
W=4
To find the length, we plug in our width.
7W-2
7(4)-2
28-2
L=26
3/6+1/6+1 2/3
Make 1 2/3 into an improper fraction:
5/3
Give them all common denominators by multiplying both sides of 5/3 by 2:
10/6
Then the question becomes:
3/6+1/6+10/6
Now you add the numerators and the denominator stays the same:
14/6
Now you can simplify by dividing both sides by 2:
7/3
Now you can make it a mixed number:
2 1/3
Hope this helps :)
Use the chain rule.
Let u = 25sin²(x), such that dy/dx = dy/du · du/dx
Answer:
As Given △QRS is rotated about point P to △Q′R′S′ .
As we know after rotation by any angle of pre-image neither the shape nor size of the image changes, the image after rotation is congruent to pre image.
Out of the options given
→The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved.
→The corresponding angle measurements in each triangle between the pre-image and the image are preserved.
are true.
