Answer:
m∠PRT = 114°
m∠T = 37°
m∠RPT = 29°
Step-by-step explanation:
This question is incomplete (without a picture) ; here is the picture attached.
In this picture, an airplane is at an altitude 12000 feet.
When the plane is at the point P, pilot can observe two towns at R and T in front of plane.
We have to find the measure of ∠PRT, ∠T and ∠RPT.
Form the figure attached segment PS is parallel to RT and PR is a transverse.
We know that internal angles formed on one side of the parallel lines by a transverse are supplementary.
Therefore, x + 66 = 180
x = 180 - 66 = 114°
∠PRT = x = 114°
m∠RPT = m∠SPR - m∠SPT
= 66 - 37
= 29°
Since m∠PRT + m∠T + m∠RPT = 180°
114 + ∠T + 29 = 180
143 + ∠T = 180
∠T = 180 - 143
∠T = 37°
Answer:
5≤u
Step-by-step explanation:
Put the words and numbers into an equation:
-5u+27≤2
Put all like terms on one side:
27-2≤5u ---> Do you see what I did there? Now you don't have to work with negatives. :)
Simplify:
25≤5u
Solve:
5≤u
:)
Answer:
1. y' = 3x² / 4y²
2. y'' = 3x/8y⁵[(4y³ – 3x³)]
Step-by-step explanation:
From the question given above, the following data were obtained:
3x³ – 4y³ = 4
y' =?
y'' =?
1. Determination of y'
To obtain y', we simply defferentiate the expression ones. This can be obtained as follow:
3x³ – 4y³ = 4
Differentiate
9x² – 12y²dy/dx = 0
Rearrange
12y²dy/dx = 9x²
Divide both side by 12y²
dy/dx = 9x² / 12y²
dy/dx = 3x² / 4y²
y' = 3x² / 4y²
2. Determination of y''
To obtain y'', we simply defferentiate above expression i.e y' = 3x² / 4y². This can be obtained as follow:
3x² / 4y²
Let:
u = 3x²
v = 4y²
Find u' and v'
u' = 6x
v' = 8ydy/dx
Applying quotient rule
y'' = [vu' – uv'] / v²
y'' = [4y²(6x) – 3x²(8ydy/dx)] / (4y²)²
y'' = [24xy² – 24x²ydy/dx] / 16y⁴
Recall:
dy/dx = 3x² / 4y²
y'' = [24xy² – 24x²y (3x² / 4y² )] / 16y⁴
y'' = [24xy² – 18x⁴/y] / 16y⁴
y'' = 1/16y⁴[24xy² – 18x⁴/y]
y'' = 1/16y⁴[(24xy³ – 18x⁴)/y]
y'' = 1/16y⁵[(24xy³ – 18x⁴)]
y'' = 6x/16y⁵[(4y³ – 3x³)]
y'' = 3x/8y⁵[(4y³ – 3x³)]
Step-by-step explanation:
remember the domain is the interval or set of valid input (x) values. the range is the interval or set of the valid result (y) values.
so, the given domain tells us the interval to look at.
what are the functional values of the function between 0 and 50 ?
since the function is continuous (there are no gaps) in this interval, we can safely assume that all values between f(0) and f(50) are valid y values.
f(0) = 4×0 + 5 = 5
f(50) = 4×50 + 5 = 205
so, the range is 5 <= y <= 205