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4 years ago

(44pts) can anyone help me ? need your help please!

Mathematics

1 answer:

Over [174]4 years ago

8 0

A. The perimeter of a right triangle is given by summing all the sides.

4x + 2 + 7x + 7 + 5x - 4

x + 3 + 2x - 5 + x + 7

b. I think you have to combine like terms of the expressions.

16x + 5

4x + 5

c.

Large triangle: 4(3) + 2 + 7(3) + 7 + 5(3) - 4 = 12+2+21+7+15-4 = 53

Small triangle: 3 + 3 + 2(3) - 5 + 3 + 7 = 3+3+6-5+3+7 = 17

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den301095 [7]

16z^2 + 25

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This happens with certain polynomials.

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Consider the curve C parametrized by x = −1 + 7 sin t and y = 1 2 − 7 cos t for −π ≤ t ≤ 5π. Then C is a circle of radius with c

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Find the general solution of the following equation. Express the solution explicitly as a function of the independent variable.

Mashcka [7]

Answer:

Therefore the general solution is

$2 \sqrt w = 2 ln(x) - 5 \frac1x +c$

Step-by-step explanation:

Integration Rule:

$\int x^n dx= \frac{x^{n+1}}{n+1}+c$
$\int \frac1x dx= ln(x) +c$

Given differential equation is

$x^2 \frac{dw}{dx}= \sqrt{w}(2x+5)$

$\Rightarrow x^2 dw= \sqrt{w} (2x+5) dx$ [ multiplying dx both sides]

$\Rightarrow \frac{dw}{\sqrt w}= \frac{(2x+5)}{x^2} dx$ [ dividing $x^2\sqrt w$ both sides]

Integrating both sides

$\int \frac{dw}{\sqrt w}=\int \frac{(2x+5)}{x^2} dx$

$\Rightarrow \int w^{-\frac12} dw=\int (\frac{2x}{x^2}+\frac{5}{x^2} )dx$

$\Rightarrow \int w^{-\frac12} dw=\int \frac{2}{x}dx +\int\frac{5}{x^2} dx$

$\Rightarrow \frac{w^{-\frac12+1}}{-\frac12+1} =2ln x+5 \frac{x^{-2+1}}{-2+1}+c$ [ c is arbitrary constant]

$\Rightarrow 2 \sqrt w = 2 ln(x) - 5 \frac1x +c$

Therefore the general solution is

$2 \sqrt w = 2 ln(x) - 5 \frac1x +c$

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3 years ago

a stained glass window is shaped like a semicircle. the bottom edge of the window is 36inches long. what is the area of the stai

lara31 [8.8K]

The formula for the area of a semicircle is equivalent to the area if it were a circle divided by 2. Since the bottom edge of the window is 36 inches long, this means that its diameter is also 36 inches long. This gives an area of 508.94 square inches.

The solution is:

A = Acircle/2 = pi*r^2/2 = pi*(18)^2/2 = 508.94 in^2

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3 years ago