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3 years ago

10 - w = 1.5 Find w.

Mathematics

2 answers:

anzhelika [568]3 years ago

6 0

Answer:

8.5

Step-by-step explanation:

w is 8.5

juin [17]3 years ago

5 0

Answer:

8.5

Step-by-step explanation:

-w=1.5-10 the opposite of 10 is negative 10

-w=-8.5

w=8.5 you are trying to get just w not negative w

so two negatives equal a positive or -1

times -8.5 is a positive 8.5

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Verify this trig function. <br> tanysiny = secy-cosy ?

avanturin [10]

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3 years ago

Find the distance from the point (1, 4) to the line y = 1/3x – 3.

Otrada [13]

Answer: Distance between line and point =

4√5 -3/2√10

Step-by-step explanation:

Distance between the line is

= √ ((9-0)²+(0+1)²)

= √ (89+1)

= √90

= 3√10

Half of the line = 3/2√10

Distance of one side of the line and the point.

= √((9-1)²+(0-4)²)

= √((8)²+(-4)²)

=√64+16

= √80

= 4√5

Distance between line and point =

4√5 -3/2√10

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3 years ago

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3 years ago

Read 2 more answers

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

The ratio of 6th grade students to 7th grade students in a soccer league is 15:10. If there are 100 students in all, how many 7t

UNO [17]

Answer:

40 7th Graders

Step-by-step explanation:

First you add all of the students to get 25. Then you divide 100 by 25 to get 4 (this is gonna be your multiplier) now multiply each part of the ration by 4, and you get 60 6th graders and 40 7th graders :D.

5 0

3 years ago

10 - w = 1.5 Find w.​

10 - w = 1.5 Find w.