^n=1/square root of 59 (535) ,if that makes sense its hard to type it
You said X = (5y+1) / (y-2)
Multiply each side by (y-2):
X(y-2) = (5y+1)
Eliminate parentheses:
Xy - 2X = 5y + 1
Add 2X to each side:
Xy = 5y + 1 + 2X
Subtract 5y from each side:
(X-5)y = 1 + 2X
Divide each side by (X-5):
<em>y = (1+2X) / (X-5) </em>
Answer:
114°
Step-by-step explanation:
The exterior angle is the sum of the remote interior angles.
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<h3>setup</h3>
(11x +15)° = 60° +6x°
<h3>solution</h3>
5x = 45 . . . . . . . . . divide by °, subtract 15+6x
x = 9 . . . . . . . . . . divide by 5
The measure of exterior angle KMN is ...
m∠KMN = (11(9) +15)° = 114°
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<em>Additional comment</em>
Both the sum of interior angles and the sum of angles of a linear pair are 180°. If M represents the interior angle at vertex M, then we have ...
60° +6x° +M = 180°
(11x +15)° +M = 180°
Equating these expressions for 180° and subtracting M gives the relation we used above:
(11x +15)° +M = 60° +6x° +M . . . . . equate the two expressions for 180°
(11x +15)° = 60° +6x° . . . . . . . . . . . subtract M
This is also described by "supplements to the same angle are equal."
Answer:
$20.60
Step-by-step explanation:
25.75x20%=5.15
25.75-5.15=20.60
The rest of the question is the attached figure.
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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