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Deffense [45]
3 years ago
11

If P and Q are fixed points, the locus of R so that <PQR=50° is:-

Mathematics
1 answer:
OleMash [197]3 years ago
3 0

Answer:

1. a line perpendicular to PQ

Step-by-step explanation:

The angle formed will be acute angle as the angle is less than 90 degrees. This line will be perpendicular to PQ . The points connected on the locus will form a semi circle. To draw the angle vertex is identified and then a line is drawn on it to form an angle.

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During gym class, Alexis played dodgeball for 34 hour.
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the soultion in 20

Step-by-step explanation:

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a road follows the shape of a parabola f(x)=3x2– 24x + 39. A road that follows the function g(x) = 3x – 15 must cross the stream
Nonamiya [84]

the coordinates where the bridges must be built is (3,-6) and (6,3) .

<u>Step-by-step explanation:</u>

Here we have , a road follows the shape of a parabola f(x)=3x2– 24x + 39. A road that follows the function g(x) = 3x – 15 must cross the stream at point A and then again at point B. Bridges must be built at those points.We need to find Identify the coordinates where the bridges must be built. Let's find out:

Basically we need to find values of x for which f(x) = g(x) :

⇒ f(x)-g(x)=0

⇒ 3x^2- 24x + 39-(3x - 15 ) =0

⇒ 3x^2- 24x + 39-3x + 15  =0

⇒ 3x^2- 27x + 54  =0

⇒ x^2- 9x + 18  =0

⇒ x^2- 6x-3x + 18  =0

⇒ x(x- 6)-3(x - 6)  =0

⇒ (x-3)(x- 6)  =0

⇒ x=3 , x=6

Value of g(x) at x = 3 : y=3x -15 = 3(3)-15 = -6

Value of g(x) at x = 6 : y=3x -15 = 3(6)-15 = 3

Therefore , the coordinates where the bridges must be built is (3,-6) and (6,3) .

7 0
3 years ago
Given that WA = 5x – 8 and WC = 3x + 2, find WB. A. WB = 5 B. WB = 8 C. WB = 10 D. WB = 17
rodikova [14]
The rest of the question is the attached figure.
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW²   → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW²  → (2)
From (1) , (2)  ⇒⇒⇒ ∴ WA = WB  →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW²   → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW²  → (5)
From (4) , (5)  ⇒⇒⇒ ∴ WC = WB  →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17

The correct answer is option D. WB = 17






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