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3 years ago

It took 4 hours to clean 10 houses, how many houses could be cleaned in 8 hours?

Mathematics

2 answers:

aleksklad [387]3 years ago

4 0

Answer:

3 1/2 houses

Step-by-step explanation:

To find the unit rate we must divide 4 by 10

4 hours divided by 10 house = 1/4

1/4 houses per hour

8 hours times 0.4= 3.2 houses

viva [34]3 years ago

4 0

Answer:

20 houses

Step-by-step explanation:

Let’s make a proportion using this setup:

houses/hours=houses/hour

We know that 10 houses can be cleaned in 4 hours. We don’t know how many houses can be cleaned in 8 hours, so we can say x houses can be cleaned in 8 hours.

10 houses/4 hours= x houses/8 hours

10/4=x/8

Now we have to solve for x, by getting x by itself. Perform the opposite of what is being done to both sides.

x is being divided by 8. The opposite of division is multiplication, so multiply both sides by 8

8(10/4)=(x/8)8

8(10/4)=x

8(2.5)=x

20=x

20 houses can be cleaned in 8 hours.

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Look at rectangle f, g, h, and j on the grid below. which rectangle has an area of 64 sqare units?

hichkok12 [17]

F= 9 x 7 = 63
G= 19 x 4 = 76
H= 12 x 4 = 48
J = 32 x 2 = 64
J is the answer

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3 years ago

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Firlakuza [10]

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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

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

3 0

3 years ago

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o-na [289]

3^3x+1 is the answer if you are simplifying

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2 years ago

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Scilla [17]

Answer:

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Step-by-step explanation:

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