Answer:
7a²/16
Step-by-step explanation:
Area of the triangle PTS
½ × a × a
a²/2
Length of PS:
sqrt(a² + a²)
asqrt(2)
Length of MS:
¼asqrt(2)
Triangles MCS and TPS are similar
With sides in the ratio:
¼asqrt(2) : a
sqrt(2)/4 : 1
Area of triangle SMC:
A/(a²/2) = [(sqrt(2)/4)]²
2A/a² = 1/8
A = a²/16
Area of PTMC
= a²/2 - a²/16
= 7a²/16
Step-by-step explanation:
A
(-2c-3d) (- 11) (- 2c-3d) (- 11) left parenthesis, minus, 2, c, minus, 3, d, right parenthesis, left parenthesis, minus, 11, right parenthesis
C
(66c + 99d) \ cdot \ dfrac {1} {3} (66c + 99d) ⋅ 3
1 left parenthesis, 66, c, plus, 99, d, right parenthesis, dot, start fraction, 1, divided by, 3, end fraction
<span> E
11\cdot(2c+3d)11⋅(2c+3d)11, dot, left parenthesis, 2, c, plus, 3, d, right parenthesis
</span> answer
(-2c-3d) (- 11) = 22c + 33d
(66c + 99d) * 1/3 = 22c + 33d
11 * ( 2c+3d) = 22c + 33d
Answer:
∂u/∂xi = i·cos(sn)
Step-by-step explanation:
For u = sin(v), the partial derivative of u with respect to xi is ...
∂u/∂xi = cos(v)·∂v/xi
In this case, v=sn, and ∂sn/∂xi = i, so the derivatives of interest are ...
∂u/∂xi = i·cos(sn)
We are given
v = 18 m/hr west
θ1 = 285°
θ2 = 340°
After 1 hours, the distance traveled by the ship is
dv = 18 mi
The distance between the ship and the lighthouse is
d = 18 / cos 340
Solve for d<span />
There are 100 centimeters in 1 meter.1 and 1/4 meters can also be written as 1.25, since 1/4 = 0.25.To find the answer, convert 1.25 meters to centimeters.
1 meter = 100 centimeters1.25 meters = x centimeters1.25 is 125% of 1. In other words, there is a 25% increase.
Now apply this increase to the centimeters side.125% = 1.251.25 • 100 = 125
So 1.25 meters (1 and 1/4 meters) equals 125 centimeters.
The string is 325 centimeters.In order for the string to be 125 centimeters, you need to cut some off.To find how much you need to cut off, subtract 125 from 325.325 - 125 = 200
Answer: In order for your string to be 1 and 1/4 meters, you must cut 200 centimeters off the original 325 centimeter long string.
