if the breadth of rectangle in one third of its length and perimeter is 80 cm find length and breadth of rectangle

Mathematics

1 answer:

LiRa [457]3 years ago

6 0

Answer:

length = 30 cm

Breadth = 10 cm

Step-by-step explanation:

Length = x cm

Breadth = (1/3)x

Perimeter = 80 cm

2(length + breadth) = 80

2* (x + (1/3)x) = 80

$2*(\frac{3}{3}x+\frac{1}{3}x) = 80\\\\2*(\frac{4}{3}x) = 80\\\\\frac{8}{3}x = 80\\\\x = 80*\frac{3}{8}\\\\x = 10*3$

x = 30cm

length = 30 cm

breadth = $\frac{1}{3}*30=10$ cm

You might be interested in

Please solve....R divided by 8

nekit [7.7K]

What divided by 8 and are there answer choices

5 0

3 years ago

Read 2 more answers

4. Find an equation for each line.

DedPeter [7]

Answer:

1 lap in 8/10 (4/5) of a minute

Answer is B, 8/10 minute!

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

Step-by-step explanation:

3 0

3 years ago

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ

Deffense [45]

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$