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2 years ago

Formula valid for all polygons: P =______________

Mathematics

2 answers:

Sergeeva-Olga [200]2 years ago

8 0

Answer:

perimeter = sum of all sides. this is same for polygons

rect: area l×b. peri = 2(l+b)

Step-by-step explanation:

15+5 = 20

20*2 = 40 cm is the perimeter of the given figure

sergiy2304 [10]2 years ago

4 0

Answer:

P = 40 cm

Step-by-step explanation:

Here,

We can see that the given polygon is a rectangle.
The length and breadth of rectangle is 15 cm and 5 cm.

As we know that the formula of perimeter of rectangle is :

$\longrightarrow{\pmb{\tt{P = 2(L + B)}}}$

P = Perimeter
L = Length
B = Breadth

Substituting all the given values in the formula to find the perimeter of rectangle :

$\begin{gathered} \qquad{\longrightarrow{\sf{P = 2(L + B)}}}\\\\\qquad{\longrightarrow{\sf{P = 2(15 + 5)}}}\\\\\qquad{\longrightarrow{\sf{P = 2(20)}}}\\\\\qquad{\longrightarrow{\sf{P = 2 \times 20}}}\\\\\qquad{\longrightarrow{\sf{P = 40}}}\\ \\ \qquad{\star{\underline{\boxed{\sf{ \red{P = 40 \: cm}}}}}}\end{gathered}$

Hence, the perimeter of rectangle is 40 cm.

$\rule{300}{2.5}$

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3 years ago

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.