Question

A wire is (7x – 3) meters long. A length of (3x – 4) meters is cut for use How much wire is left?​

Answer 1

Hello, Buddy!!

||Required Solution||

Given,

Length of wire = (7x-3) meters.

Length of wire used = (3x-4) metres

Let, Remaining wire be y

→ y+(3x-4) = (7x-3)

→ y = (7x-3)-(3x-4)

→ y = 7x-3-3x+4

→ y = 4x+1

• Length of Wire Left ☞ (4x+1) meters.

@MrMonarque♡

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Answer 2

$\large\boxed{\textsf{\textbf{\red{Question\::-}}}}$

• A wire is (7x−3) meter long. A length of (3x−4) meters is cut for use. Now, answer the following questions:
• (a) How much wire is left?
• (b) If this left out wire is used for making an equilateral triangle. What is the length of each side of the triangle so formed?

$\large\boxed{\textsf{\textbf{\red{Answer\::-}}}}$

(a) Given,

Length of a wire = (7x-3)

and wire cut for use has length =(3x−4)m

Left wire =(7x−3)−(3x−4)=(4x+1)m

(b) Left wire =(4x+1)m

Perimeter of equilateral triangle = length of wire left

$⇒3 \times \: side = 4x + 1$

$⇒side = \frac{(4x + 1)}{3} = \frac{1}{3}(4x + 1) m$