The ΔABR ≅ ΔACR are congruent by SAS theorem
Option D is the correct answer.
The missing diagram is attched with the answer.
<h3>What is a Triangle ?</h3>
A triangle is a polygon with three sides , three vertices and three angles.
SAS will be used to prove the congruence of ΔABR ≅ ΔACR
In both the triangle we have a common side , AR
AB = AC (given)
∠ B A R = ∠R A C ( given equal)
So as the two side and the included angle is equal
Therefore the ΔABR ≅ ΔACR are congruent by SAS theorem
Option D is the correct answer.
To know more about Triangle
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Since Ryan is 5 feet tall, we will assume that the height of the cell-phone tower is x + 5.
Since the given is 54 degrees.
Tan 54degrees = x/80(feet)
x=80(tan54)
x=110
To finalize the answer, we will add up the height of the cell-phone tower to the answer.
110 + 5 = 115 feet
So the answer to the question is that the height of the cell-phone tower is 115 ft.
Answer:
3 years 8 month
Step-by-step explanation:
Given 7 years 6 months - 3 years 10 months
Step 1: Firstly we should subtract months from months then after from year to year.
Step 2: 6 months is less than 10 months so we have to borrow 1 year and add it to the months.
Step 3: Adding 1 year to 6 months we get 18 months.
Step 4: Now we have to subtract 3 years 10 months from 6 years 18 months.
Step 5: After subtracting we get 3 years 8 months.
Thus the answer is 3 years 8 months.
Answer:
"the sum of the cube of 6x and 2"
Step-by-step explanation:
We see that only the expression of 6x is being cubed, not the entire expression of (6x)³ + 2.
The first blank must apply to both these terms 6x and 2, so the word can't be cubed. Instead, because we're finding the sum of these two, the word should be "sum".
The second blank should apply to only the 6x because we've already described the relationship of 6x and 2. So, since we are cubing 6x, the word here should be "cube".
The final sentence should read:
"the sum of the cube of 6x and 2"
Answer:
a. v(t)= -6.78
+ 16.33 b. 16.33 m/s
Step-by-step explanation:
The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=
=
. We now multiply both sides of the equation by the integrating factor.
μv' + μkv = μg ⇒ v' + kv = g ⇒ [v]' = g. Integrating, we have
∫ [v]' = ∫g
v = 
+ c
v(t)= + c
.
From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have
9.55 = 9.8 × 15/9 + c
= 16.33 + c
c = 9.55 -16.33 = -6.78.
So, v(t)= 16.33 - 6.78. m/s = - 6.78 + 16.33 m/s
b. Velocity of object at time t = 0.5
At t = 0.5, v = - 6.78
+ 16.33 m/s = 16.328 m/s ≅ 16.33 m/s
