Written form is just how you would usually see a number. So it would be 3,680,010.
B,C and D are correct because all of those choices include letters which are arranged in order from left of right.
9514 1404 393
Answer:
- tan(A) = 7/24
- sin(A) = 7/25
- cos(A) = 24/25
Step-by-step explanation:
The mnemonic SOH CAH TOA can help you remember the trig ratios:
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
Tan = Opposite/Adjacent
With respect to angle A, the sides are Adjacent = 24; Opposite = 7, Hypotenuse = 25. Then the desired trig ratios are ...
tan(A) = 7/24
sin(A) = 7/25
cos(A) = 24/25
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
Answer:
Part a) The radii are segments AC and AD and the tangents are the segments CE and DE
Part b) 
Step-by-step explanation:
Part a)
we know that
A <u>radius</u> is a line from any point on the circumference to the center of the circle
A <u>tangent</u> to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.
In this problem
The radii are the segments AC and AD
The tangents are the segments CE and DE
Part b)
we know that
radius AC is perpendicular to the tangent CE
radius AD is perpendicular to the tangent DE
CE=DE
Triangle ACE is congruent with triangle ADE
Applying the Pythagoras Theorem
substitute the values and solve for CE
remember that
CE=DE
so
