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2 years ago

Help me please, im struggling

Mathematics

2 answers:

vagabundo [1.1K]2 years ago

6 0

Answer:

8.57

Step-by-step explanation:

Hope it helps.......

Elena L [17]2 years ago

4 0

Answer:

x = 8.57

Step-by-step explanation:

Hi! So first of all, I attached an image that has my best visual explanation of how the 2 triangles overlap but are indeed two separate triangles!

Since the question states that the triangles are similar, that basically means in simple terms that their sides are proportional or, are all based on the same fraction (proportion)! (An example of this is the triangles with the sides that correlate [in this problem, 11 to 7] would be 1 to 2, 3 to 6, and 4 to 8)

Now, we have to figure out which of our sides correspond, the first one being 11 to 7, and the second one being x + 15 (which I got by adding the small portion I have from the big triangle, and the portion I have from the smaller triangle together because the small triangle is inside of the big one!) to 15.

Next, I have to find out what x is based on the fraction (proportion) I already have which is 11 to 7 or $\frac{11}{7}$ . Then, I compare this by setting this proportion equal to are other proportion with x in it!

$\frac{11}{7} = \frac{x+15}{15}$

Finally, I have to get x by itself to find what it is equal to, which you can see in the image! (If you have a number in the denominator and a variable in the numerator, and a regular number or fraction without a variable on the other side, you can just multiply both sides by the number in the denominator to cancel out the bottom of said variable fraction!)

So! X is equal to 8.57!

Hope this Helps! :)

Have any questions? Ask below in the comments and I will try my best to answer.

-SGO

#SPJ1

4 0

1 year ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

Find the directional derivative of the function at the given point in the direction of the vector v. G(r, s) = tan−1(rs), (1, 3)

alexandr1967 [171]

The directional derivative of $f$ at the given point in the direction indicated is $\frac{5}{2}$ .

<h3>How to calculate the directional derivative of a multivariate function</h3>

The directional derivative is represented by the following formula:

$\nabla_{\vec v} f = \nabla f (r_{o}, s_{o})\cdot \vec v$ (1)

Where:

$\nabla f (r_{o}, s_{o})$ - Gradient evaluated at the point $(r_{o}, s_{o})$ .
$\vec v$ - Directional vector.

The gradient of $f$ is calculated below:

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial r}(r_{o},s_{o}) \\\frac{\partial f}{\partial s}(r_{o},s_{o}) \end{array}\right]$ (2)

Where $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial s}$ are the partial derivatives with respect to $r$ and $s$ , respectively.

If we know that $(r_{o}, s_{o}) = (1, 3)$ , then the gradient is:

$\nabla f(r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{s}{1+r^{2}\cdot s^{2}} \\\frac{r}{1+r^{2}\cdot s^{2}}\end{array}\right]$

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{1+1^{2}\cdot 3^{2}} \\\frac{1}{1+1^{2}\cdot 3^{2}} \end{array}\right]$

$\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right]$

If we know that $\vec v = 5\,\hat{i} + 10\,\hat{j}$ , then the directional derivative is:

$\nabla_{\vec v} f = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right] \cdot \left[\begin{array}{cc}5\\10\end{array}\right]$

$\nabla _{\vec v} f (r_{o}, s_{o}) = \frac{5}{2}$

The directional derivative of $f$ at the given point in the direction indicated is $\frac{5}{2}$ . $\blacksquare$

To learn more on directional derivative, we kindly invite to check this verified question: brainly.com/question/9964491

3 0

2 years ago