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GarryVolchara [31]

2 years ago

12

En la resolución de problemas para verificar la respuesta se debe de tener en cuenta el resultado y los datos desconocidos verda

dero o falso

1 answer:

kvv77 [185]2 years ago

7 0

Answer:ha

Step-by-step explanation:

I did something

You might be interested in

Problem PageQuestion The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate paramet

asambeis [7]

Answer:

<em>t = 1.51</em>

Step-by-step explanation:

<u>Exponential Model</u>

The exponential model is often used to simulate the behavior of a magnitude that either grow or decay in proportion to the existing amount of that magnitude.

The model can be expressed as

$M=M_oe^{kt}$

In this case, Mo is the initial mass of the radioactive substance and k is a constant which value is positive if the mass is growing or negative if the mass is decaying.

The value of k is not precisely given in the question, we are assuming $k=-0.2$

The model is now

$M=M_oe^{-0.2t}$

We are required to compute the time it takes the mass to reach one-half of its initial value:

$\displaystyle \frac{M_o}{2}=M_oe^{-0.2t}$

Simplifying

$\displaystyle \frac{1}{2}=e^{-0.2t}$

Taking logarithms

$\displaystyle ln\frac{1}{2}=ln(e^{-0.2t})=-0.2t$

Solving for t

$\displaystyle t=-\frac{ln\frac{1}{2}}{0.2}=1.51$

6 0

3 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago

Why are we teching hligh levels of math at lower grades

Oksanka [162]

Answer:

I have no idea

Step-by-step explanation:

people are weird

8 0

2 years ago

Which equation demonstrates the distributive property?

PIT_PIT [208]

B because distributive property goes like this:
a(b+c)
a(b)+a(c)
That's just an example though

4 0

3 years ago

Read 2 more answers

Can someone please help me with a few of theseee

charle [14.2K]

i dont know, sorry.

7 0

2 years ago