All categories

3 years ago

What's 5 - the square root of x + 10 = the square root of 7 - x

1 answer:

3 0

$\bf 5-\sqrt{x+10}=\sqrt{7-x}\impliedby \textit{squaring both sides}
\\\\\\
(5-\sqrt{x+10})^2=(\sqrt{7-x})^2
\\\\\\
(5^2)-10\sqrt{x+10}+(x+10)=7-x
\\\\\\
35-10\sqrt{x+10}+x=7-x\implies 28-10\sqrt{x+10}+2x=0
\\\\\\
2x+28=10\sqrt{x+10}\implies x+14=5\sqrt{x+10}\impliedby 
\begin{array}{llll}
squaring\ both\\
sides\ again
\end{array}
\\\\\\
$

$\bf (x^2)+28x+(14^2)=5^2(x+10)\implies x^2+28x+196=25x+250
\\\\\\
\begin{array}{lcclll}
x^2&+3x&-54&=0\\
&\uparrow &\uparrow \\
&-6+9&-6\cdot 9
\end{array} \implies (x+9)(x-6)=0\implies 
\begin{cases}
x+9=0\\
\boxed{x=-9}\\
-----\\
x-6=0\\
\boxed{x=6}
\end{cases}$

You might be interested in

HELP ASAP Translate 6(4j+5+4j) in to a verbal expression w step by step. WILL MARK BRAINLIEST

natali 33 [55]

The answer would be 24j+30+24j which is 48j+30

8 0

2 years ago

Consider the ODE, dy dx = y 2 1 + x (2) subject to condition y = 1 when x = 0, use your Euler code from class (modified if neces

makkiz [27]

Answer:

Computation.

Step-by-step explanation:

I'm not really sure if that's the analytical solution of the inital value problem,

because y(0)=11-ln(1-0)(3)=11. Howevwer, let us procede with the given values...

Let us assume that we are going to use euler with n=2 (two steps) and h=0.2(the size of each step)

The update rules of the Euler Methode are

$X_i = X_{i-1}+h=X_0+ih$

$m_i=\dfrac{dY}{dX} \biggr \rvert_{x_i} \\\\Y_{i+1}=m_i\cdot h+y_i$

Since the initial value problem tells us that Y=1 when X=0, we know that

$X_0=0$ and that $Y_0=1$ . Then, we have

$X_0=0\\\\X_1=0.2\\\\X_2=0.4$

and

$Y_0=1\\\\m_0=21 \cdot Y_0 + X_0 \cdot 2=21 \cdot 1 + 0=21\\\\Y_1=0.2 \cdot 21 + 1 =5.2\\\\m_1=21 \cdot Y_1 + X_1\cdot 2=21 \cdot 5.2 + 0.2 \cdot 2=109.6\\ \\Y_2=m_1 \cdot h + Y_1 = 109.6 \cdot 0.2 + 5.2= 27.12$

which gives us the points (0,1), (0.2, 5.2) and (0.4, 27.12).

Now, since we want to compare the analyticaland the Euler result, we first compute the value of y=11-ln(1-x)(3) for the values x=0, 0.2 and 0.4. We get that

$y(0)=11-\ln(1-0)(3)=11-ln(1)(3)=11\\\\Y(1)=11-\ln(1-0.2)(3)=11.67\\\\Y(2)=11-\ln(1-0.4)(3)=12.53$

and we compute $Y(i)-Y_i$ for each i.

It holds

$Y(0)-Y_0=11-1=10\\\\Y(1)-Y_1=11.67-5.2=6.47\\\\Y(2)-Y_2=12.53-27.12=-14.59$

which tells us that we have a really bad approximation, as I already stated there must be a mistake in the analytical solution since the intial values don't coincide. Also note that the curve that we get using the euler methose is growing faster than the analitical solution.

4 0

2 years ago

Jada was using square stickers with a side length of 3/434 inch to decorate the spine of a photo album. The spine is 10 1/21012

Yuki888 [10]

Answer: The number of stickers she used to cover the length of the spine = 14.

Step-by-step explanation:

Given: The length of spine = $10\dfrac12$ inches = $\dfrac{21}{2}$ inches

Length of each sticker = $\dfrac34$ inch

If she laid the stickers side by side without gaps or overlaps, the number of stickers she use to cover the length of the spine = $\text{(Length of spine)}\div (\text{Length of each sticker})$