<span>Since the triangles are right triangles, the hypotenuse can be calculated using the Pythagorean theorem:
5²+12²=x²
x²=25+144, x²=169⇒x=√169
x=13
The hypotenuse of a smaller triangle is 13 units.</span>

8 0

3 years ago

Read 2 more answers

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago

A square base that covers an area of 25 ft squared what is the length of one side of the tent?

sukhopar [10]

I was also going to say 6.25 but someone else did :)

5 0

3 years ago

3x+2(5x-3)=7<br> with steps

EleoNora [17]

Answer: x=1

Step-by-step explanation: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3*x+2*(5*x-3)-(7)=0

3x + 2 • (5x - 3)) - 7 = 0

Pull out like factors :

13x - 13 = 13 • (x - 1)

13 = 0

A a non-zero constant never equals zero.

Solve : x-1 = 0

Add 1 to both sides of the equation :

x = 1

6 0

3 years ago