Answer:
B:rate
Step-by-step explanation:
the formula distance= rate × time
Answer:
80p + 48q - 8
Step-by-step explanation:
Hello, here are the fraction forms for this equation
Irrational fraction form; -153/20
Mixed number form; -7 13/20
Hoped this helped
Answer:
S = (-4, -34)
Step-by-step explanation:
T=(10, 18) is one end of the segment
the midpoint is located at (5, -8)
therefore, using the formula for midpoint, we can find the other end of the segment (point S)
Midpoint x-value = xa + (xb - xa)/2
in our case;
5 = 10 + (xb-10)/2
3 - 10 = (xb-10)/2
-7 = (xb-10)/2
-14 = xb-10
xb = -14 + 10 = - 4
Now for the y value of the midpoint:
Midpoint y-value = ya + (yb - ya)/2
in our case:
-8 = 18 + (yb - 18)/2
- 26 = (yb - 18)/2
- 52 = yb - 18
yb = -52 + 18 = -34
Then. point S is located at: S = (-4, -34)
Answer:
ABC shaded area = 36
- 72 cm²
ABC shaded area perimeter =
cm
ABCD area =
cm²
ABCD perimeter =
cm
Step-by-step explanation:
<u>Shape ABC</u>
Assuming you want the area and perimeter of the shaded part of the shape only...
<u>Area</u>
Area of a sector =
(where r is the radius and
<em> </em>
⇒ area of a sector = ![\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2](https://tex.z-dn.net/?f=%5Cdfrac12%20%5Ctimes%2012%5E2%5Ctimes%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%20%3D36%5Cpi%20%20%5C%20%5Ctextsf%7Bcm%7D%5E2)
Area of triangle = 1/2 x base x height
⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²
Therefore, area of shaded area = area of sector - area of triangle
⇒ area = 36
- 72 cm²
<u>Perimeter</u>
Arc length =
(where r is the radius and
<em> </em>
⇒ arc length = ![12\times\dfrac12\pi =6\pi \ \textsf{cm}](https://tex.z-dn.net/?f=12%5Ctimes%5Cdfrac12%5Cpi%20%3D6%5Cpi%20%20%5C%20%5Ctextsf%7Bcm%7D)
Hypotenuse of triangle =
(where a and b are the legs of the right triangle)
⇒ hypotenuse =
cm
Therefore, perimeter = arc length + hypotenuse
⇒ perimeter =
cm
<u>Shape ABCD</u>
<u>Area</u>
Area of a semicircle =
(where r is the radius)
⇒ area of large semicircle ABC = ![\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2](https://tex.z-dn.net/?f=%5Cdfrac12%20%5Ctimes%20%5Cpi%20%5Ctimes%202%5E2%3D2%5Cpi%20%20%5C%20%5Ctextsf%7Bcm%7D%5E2)
⇒ area of small semicircle AD = ![\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2](https://tex.z-dn.net/?f=%5Cdfrac12%20%5Ctimes%20%5Cpi%20%5Ctimes%201%5E2%3D%5Cdfrac12%5Cpi%20%20%5C%20%5Ctextsf%7Bcm%7D%5E2)
⇒ area of shape ABCD = ![\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2](https://tex.z-dn.net/?f=%5Cdfrac12%20%5Cpi%20%2B%202%20%5Cpi%3D%5Cdfrac52%20%5Cpi%20%5C%20%5Ctextsf%7Bcm%7D%5E2)
<u>Perimeter</u>
1/2 circumference = ![\pi r](https://tex.z-dn.net/?f=%5Cpi%20r)
⇒ perimeter = ![2\pi +2+\pi=3 \pi+2 \ \textsf{cm}](https://tex.z-dn.net/?f=2%5Cpi%20%2B2%2B%5Cpi%3D3%20%5Cpi%2B2%20%5C%20%5Ctextsf%7Bcm%7D)