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3 years ago

PLEASE HELP !! WILL GIVE BRAINLIEST 20 POINTS

Mathematics

1 answer:

REY [17]3 years ago

5 0

Answer:

use desmos to figure it out.

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A new sweater costs $15.99. If the sweater is on sale for 1/4 off

Mrac [35]

the sweater costs $15.99 regularly, however today is on sale, 1/4 off the regular price, how much is 1/4 of 15.99? well just their product, 15.99 * (1/4) = 3.9975.

that means that just for today, the sweater costs 15.99 - 3.9975.

so, today you're not really paying $15.99 for the sweater, you're paying 3.9975 less, so you're saving 3.9975. That's $3.9975 that you won't be spending on it, thus saving it.

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4 years ago

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Translate the sentence into an equation.

posledela

The equation would be 2x+8=7

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3 years ago

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What is the quotient of 5 1/4 divided by 15

liberstina [14]

.35 is your answer that your looking for

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3 years ago

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Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7

Ilia_Sergeevich [38]

I'll go out on a limb and guess the system is

$\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x$

with initial condition $\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top$ . The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12$

The corresponding eigenvectors $\eta$ are such that

$\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}$

$\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}$

So the characteristic solution to the ODE system is

$\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

When $t=0$ , we have

$\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}$

from which it follows that $C_1=2$ and $C_2=5$ , making the particular solution to the IVP

$\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

$\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}$

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4 years ago

What's 20000 times a million thousand

shutvik [7]

20000 times a million thousand it equal 20000000000

4 0

3 years ago