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Darina [25.2K]
3 years ago
7

4. The distance between a + 7i and -8 + 4i

Mathematics
1 answer:
Elis [28]3 years ago
4 0

Answer:

<h2>a = 3</h2>

Step-by-step explanation:

\text{The formula of a distance between two complex numbers}\\\\z_1=a+bi,\ z_2=c+di:\\\\d(z_1,\ z_2)=|z_2-z_1|\ \text{where}\ |z|=\sqrt{Re^2+Im^2}\\\\=============================

z_1=a+7i,\ z_2=-8+4i\\\\d(z_1,\ z_2)=|(-8+4i)-(a+7i)|=|-8+4i-a-7i|=|-8-a-3i|\\\\=\sqrt{(-8-a)^2+3^2}\\\\d(z_1,\ z_2)=\sqrt{130},\\\\\text{therefore}\ \sqrt{(-8-a)^2+3^2}=\sqrt{130}\Rightarrow(-8-a)^2+3^2=130\\\\\bigg(-(8+a)\bigg)^2+9=130\qquad\text{subtract 9 from both sides}\\\\(8+a)^2=121\iff8+a=\pm\sqrt{121}\\\\8+a=\pm11\qquad\text{subtract 8 from both sides}\\\\a=-19\ or\ a=3

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Given that WA = 5x – 8 and WC = 3x + 2, find WB. A. WB = 5 B. WB = 8 C. WB = 10 D. WB = 17
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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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-14x>-3

+14  +14

x > 11

6 0
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