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2 years ago

The areas of the squares adjacent to two sides of a right triangle are 32units^2 and 32 units^2

Mathematics

2 answers:

natima [27]2 years ago

8 0

Answer:

Step-by-step explanationmoo

adoni [48]2 years ago

4 0

Answer:

x = 8 units.

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

<em>The areas of the squares adjacent to two sides of a right triangle are 32 unit square and 32 units square. Find the length,x, of the third side of the triangle</em>

My answer:

Let a is the side length of the 1st square

The area of the 1st square is equal to 32 units square

<=> $a^{2} = 32$

<=> a = $4\sqrt{2}$ units

Let b is the length of the 2nd square

The area of the 2nd square is equal to 32 units square

<=> $b^{2}$ = 32

<=> b = $4\sqrt{2}$ units

Find the value of x

Applying the Pythagoras Theorem, we have:

$x^{2} = a^{2} + b^{2}$

<=> $x^{2} = 32 + 32$

<=> $x^{2} = 64$

<=> x = 8 units

Hope it will find you well.

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vova2212 [387]

Answer:

Step-by-step explanation:

surface area = 2πrh + 2πr²

where r = 2 miles radius

h = 3 miles

<u>plugin values into the formula</u>

surface area = 2πrh + 2πr²

= 2π (2) 3 + 2π (2)²

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= 62.832 mi.²

4 0

3 years ago

Write an equation for the line in point slope form through the points (2, −6) and (−1, −8).

igomit [66]

Let point A be $(2, -6)$ and point B $(-1, -8)$

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-8 - (-6)}{-1 - 2} = \frac{-8 + 6}{-1 -2} = \frac{-2}{-3} = \frac{2}{3}$

Use the point-slope formula and whichever point you want as , in this case I'll use

$y - y_1 = m(x - x_1)\\y - (-6) = \frac{2}{3}(x - 2)\\y + 6 = \frac{2}{3}x - \frac{4}{3}\\y = \frac{2}{3}x - \frac{4}{3} - 6\\y = \frac{2}{3}x - \frac{14}{3}$

8 0

2 years ago

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

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