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2 years ago

I need some help with this math equation please lol

Mathematics

1 answer:

Nimfa-mama [501]2 years ago

4 0

Answer:

3/6 or 1/2 simplified

Step-by-step explanation:

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Please help! giving brainliest.

Advocard [28]

Answer:

99.225

Step-by-step explanation:

10×3.15 is 31.5

31.5 squared is 99.225

Hope this helps!

7 0

2 years ago

Read 2 more answers

PLZ HELP 75 POINTS SHOW WORK

Alexeev081 [22]

Answer:

See in attachment and Mark as brainleist please.

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2 years ago

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Mrs Harvey is making fresh raspberry punch. Each pitcher needs 1/2 cup of raspberries. If she makes three pitcher how many cups

Nastasia [14]

Answer:

1.5 cups of raspberries

Step-by-step explanation:

1/2×3 equals 1.5

7 0

2 years ago

#4 home do I find which one it is?

lina2011 [118]

$\bf \begin{array}{lllll}
solutions&graphs&slopes\\
----&----&----\\
\textit{exactly one}&
\begin{array}{llll}
\textit{the two lines intersect}\\
\textit{at one point}
\end{array}&\textit{different slopes}\\\\
infinitely\quad many&
\begin{array}{llll}
\textit{the two lines coincide}\\
\textit{one is right on top}\\
\textit{ of the other}
\end{array}&
\begin{array}{llll}
\textit{equal slopes}\\
\textit{equal y-intercepts}
\end{array}
\end{array}$

$\bf \textit{no solution}\qquad\quad &\textit{lines are parallel} \qquad &
\begin{array}{llll}
\textit{equal slopes}\\
\textit{different y-intercepts}
\end{array}
\end{array}$

for example, let's look at the first set

y+3x =5 or y = -3x+ 5
and y = -3x + 2
y = m + b

the slopes are equal, the y-intercepts differ
that means, they're just parallel lines, no solution

8 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago