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2 years ago

What is the perimeter of this polygon? Round your answer to the nearest tenth

Mathematics

2 answers:

bulgar [2K]2 years ago

6 0

Answer:

Step-by-step explanation:

pythagorean theorem

A^2+B^2=C^2 do that for all sides then add

Degger [83]2 years ago

4 0

Answer:

A) 21.6 units

Step-by-step explanation:

Use the distance formula for each length:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\d=\sqrt{(-4-2)^2+(0-3)^2}\\d=\sqrt{36+9}\\d=\sqrt{45}\\\\ d=\sqrt{(2-4)^2+(3-0)^2}\\d=\sqrt{4+9}\\d=\sqrt{13}\\\\d=\sqrt{(4-0)^2+(0--4)^2}\\d=\sqrt{16+16}\\d=\sqrt{32}\\\\d=\sqrt{(0--4)^2+(-4-0)^2}\\d=\sqrt{16+16}\\d=\sqrt{32}\\\\p=\sqrt{45}+\sqrt{13}+\sqrt{32}+\sqrt{32}=21.62746371$

This means that option A is the closest answer

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Order of operations (GEMS)

cestrela7 [59]

(6+2*3)/(9-7)^2
(6+6)/(2)^2
12/4
3

Final answer: 3
Remember to simplify within parentheses first and then divide the two values in the end.

3 0

3 years ago

Area of the trapezoid 1/2 3ft 5 ft 2 ft

zzz [600]

Answer:

3 ft

Step-by-step explanation:

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2 years ago

How many 5/8s are in 3

In-s [12.5K]

Well, in 3 there are 24/8s.
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2 years ago

Read 2 more answers

What is the least common denominator of the fractions 1/3 5/6 5/9

hichkok12 [17]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago