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2 years ago

A wood stove burns 4 logs in 2 hours how many logs burn in 9 hours ?

Mathematics

2 answers:

ziro4ka [17]2 years ago

8 0

Answer:

18 logs.

Step-by-step explanation:

Since a stove burns 4 logs per 2 hours, then the stove will burn 2 logs in one hour. There fore in 9 hours the stove will burn 2*9 which is 18 logs.

4:2= 2 logs per hour

2*9=18 logs per 9 hours.

Sever21 [200]2 years ago

5 0

Answer:

Step-by-step explanation:

16 Because 2 divided by 8 equals 4 then 4x4 equal 16

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Can someone help with these two questions? dont worry if its wrong!

nevsk [136]

Answer:

1)8&3/4 CUP FLOUR

2)26 KIDS CAN GO ..

Step-by-step explanation:

1) one batch=2 1/2 cup flour

therefore batch required in 3 1/2 batch= 5/2×7/2=35/4

=8&3/4cup flour

2)No. of kids can go for $60=8

therefore,no. of kids can go in $1=8/60

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2 years ago

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Carlos finished 1/3 of his art project on Monday. Tyler finished 1/2 of his art project on Monday. Who finished more of hood art

vladimir2022 [97]

Tyler finished more of the art project.

Carlos finished 1/3
In terms of percentage 1/3 *100% = 33.3%

Tyler finished 1/2
In terms of percentage 1/2 *100% = 50%

50% > 33.3%
50% is greater than 33.3%

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2 years ago

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5m - 8 > 12 solve each problem showing your steps

Mashutka [201]

Answer:

m > 4

Step-by-step explanation:

5m - 8 > 12

Add 8 to each side

5m - 8+8 > 12+8

5m > 20

Divide by 5

5m/5 > 20/5

m > 4

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2 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

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

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3 years ago

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Ksivusya [100]

Answer:

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