Answer:
A) AAS; B) LA; C) ASA
Step-by-step explanation:
AAS is the Angle-Angle-Side congruence statement. It says that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of a second triangle, then the triangles are congruent. In these triangles, ∠E≅∠K, ∠F≅∠L, and DE≅JK. These are two angles and a non-included side; this is AAS.
LA is the leg-acute theorem. It states that if a leg and acute angle of one triangle is congruent to the corresponding leg and acute angle of another triangle, then the triangles are congruent.
The leg we have congruent from each triangle is DE and JK. We also have ∠E≅∠K and ∠F≅∠L, both pairs of which are acute. This is the LA theorem.
ASA is the Angle-Side-Angle congruence statement. It says that if two angles and an included side of one triangle are congruent to the corresponding two angles and included side of another triangle, then the triangles are congruent.
We have that ∠D≅∠J, DE≅JK and ∠E≅∠K. This gives us two angles and an included side, or ASA.
To get the that relates x and y, we choose two points on the table and use the equation of a straight line as follows:
Recall that the equation of a staight line is obtained from the formula
Using the points (30, 18) and (36, 24), we have
Answer:
the answer will be 18 ohkay but they will eill have more dimes than quarters
Answer:
ln|sec θ + tan θ| + C
Step-by-step explanation:
The integrals of basic trig functions are:
∫ sin θ dθ = -cos θ + C
∫ cos θ dθ = sin θ + C
∫ csc θ dθ = -ln|csc θ + cot θ| + C
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ tan θ dθ = -ln|cos θ| + C
∫ cot θ dθ = ln|sin θ| + C
The integral of sec θ can be proven by multiplying and dividing by sec θ + tan θ, then using ∫ du/u = ln|u| + C.
∫ sec θ dθ
∫ sec θ (sec θ + tan θ) / (sec θ + tan θ) dθ
∫ (sec² θ + sec θ tan θ) / (sec θ + tan θ) dθ
ln|sec θ + tan θ| + C
Answer:
R (t) = 60 - 60 cos (6t)
Step-by-step explanation:
Given that:
R(t) = acos (bt) + d
at t= 0
R(0) = 0
0 = acos (0) + d
a + d = 0 ----- (1)
After 
seconds it reaches a height of 60 cm from the ground.
i.e
Recall from the question that:
At t = 0, R(0) = 0 which is the minimum
as such it is only when a is negative can acos (bt ) + d can get to minimum at t= 0
Similarly; 60 × 2 = maximum
R'(t) = -ab sin (bt) =0
bt = k π
here;
k is the integer
making t the subject of the formula, we have:
replacing the derived equation of k into R(t) = acos (bt) + d
Since we known a < 0 (negative)
then d-a will be maximum
d-a = 60 × 2
d-a = 120 ----- (3)
Relating to equation (1) and (3)
a = -60 and d = 60
∴ R(t) = 60 - 60 cos (bt)
Similarly;
For 
where ;
Then b = 6
∴
R (t) = 60 - 60 cos (6t)
