Answer:
r = 12
Step-by-step explanation:
From the figure attached,
QP is a tangent to the circle O at the point P.
Therefore, by the property of tangency,
OP ⊥ QP
By applying Pythagoras theorem In right triangle QPO,
(Hypotenuse)² = (Leg 1)² + (Leg 2)²
(OQ)² = (OP)² + (PQ)²
(25 + r)² = (35)² + r²
625 + r² + 50r = 1225 + r²
50r = 1225 - 625
50r = 600
r = 12
Therefore, r = 12 units is the answer.
The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
</u>
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.
Answer:
The answer is Tangent.
Step-by-step explanation:
Given that Tangent Rule is tanθ = opposite/adjacent. In this question, opposite is 11, adjacent is x and θ is 21°.
So using this formula, you are able to find the value of x.
This is a disjunction, B, because it has 2 statements with an OR in the middle.