The correct answer is B). 6
Radius of inscribed circle of triangle HJK is 6 unit
Step-by-step explanation:
As shown in figure,
The triangle HJK has In-circle or inscribed circle of radius PL(or PM or PN)
Since, PL and PM is radius of same in-circle,
PL = PM
Given that PL=3x+4 and PM =6x-14
3x+4=6x-14
3x=18
x=6
Thus, The correct answer is B). 6
Answer:
$2,459.21
Step-by-step explanation:
(see attached for reference)
recall that the formula for compound interest is:
A = P [ 1 + (r/n)^ (nt) ]
where,
A = Final amount ( we are asked to find this)
P = principal amount = given as $2,340
r = Annual Interest Rate = given as 5% = 0.05
n = number of times compounded in a year = 4 (compounded quarterly)
t= time = 1 year
Substituting the values into the equation,
A = P [ 1 + (r/n)^ (nt) ]
A = 2,340 [ 1 + (0.05/4)^ (4·1) ]
A = $2,459.21
A, because anything with a dilation will not be congruent as the original PQR :))
point slope form:y-2=4(x+1)
then distribute 4 and simplify y-2=4x+4 , y=4x+6 is the slope -y intercept form
then, move 4x on the left side, you get -4x+y=6 , multiply every term by -1, you get 4x-y=-6 which is standard form.
Answer:
Correct integral, third graph
Step-by-step explanation:
Assuming that your answer was 'tan³(θ)/3 + C,' you have the right integral. We would have to solve for the integral using u-substitution. Let's start.
Given : ∫ tan²(θ)sec²(θ)dθ
Applying u-substitution : u = tan(θ),
=> ∫ u²du
Apply the power rule ' ∫ xᵃdx = x^(a+1)/a+1 ' : u^(2+1)/ 2+1
Substitute back u = tan(θ) : tan^2+1(θ)/2+1
Simplify : 1/3tan³(θ)
Hence the integral ' ∫ tan²(θ)sec²(θ)dθ ' = ' 1/3tan³(θ). ' Your solution was rewritten in a different format, but it was the same answer. Now let's move on to the graphing portion. The attachment represents F(θ). f(θ) is an upward facing parabola, so your graph will be the third one.