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3 years ago

I will give brainliest if correct!!! Please show work so I know how to do it. :)

Mathematics

2 answers:

yawa3891 [41]3 years ago

5 0

Answer:

.2625

Step-by-step explanation:

logb(3) = 0.5646, logb(4) = 0.7124, and logb(5) = 0.8271

logb(5/3)

We know that logx ( y/z) = logx (y) - logz (z)

logb ( 5) - logb(3)

0.8271 - 0.5646

.2625

kramer3 years ago

3 0

Answer:

F. 2.6252

Step-by-step explanation:

0.8271 - 0.5646

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What is 80% of 960? Round to the nearest hundredth (if necessary).

Tatiana [17]

Answer:

Step-by-step explanation:

you could do (960/100)*80=768

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3 years ago

CAN SOMEONE ANSWER ASAP PLEASE!!!!! WILL GIVE EXTRA POINTS

sdas [7]

I'm really sorry if I'm wrong but

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Step 2:
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3 0

3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

4 years ago

Add or subtract as indicated 5/x-2 - 3/x-2

algol [13]

Option a: $\frac{2}{x-2}$ is the right answer

Step-by-step explanation:

The given expression is:

$\frac{5}{x-2} - \frac{3}{x-2}$

As the denominators of both terms are same, the LCM will be x-2

$\frac{5-3}{x-2}$

Subtracting

$\frac{2}{x-2}$

Hence,

Option a: $\frac{2}{x-2}$ is the right answer

Keywords: Polynomials, fractions

Learn more about polynomials at:

#LearnwithBrainly

4 0

3 years ago