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3 years ago

If you help me get this right i will mark u as most brilliance!

Mathematics

2 answers:

Gekata [30.6K]3 years ago

7 0

Answer:

31.7

Step-by-step explanation:

bro literally just type 4 sqrt 63 into google. smh. its 31.7 tho

ExtremeBDS [4]3 years ago

7 0

Answer:

31.7

Step-by-step explanation:

I used a calculator hope that helps

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составь верные равенства и неравенства, используя таблицу: 32×2, 120+253, 900-653, 300×2, 640÷8, 900÷3, 450-350, 45×3

tekilochka [14]

What is your question?

5 0

3 years ago

Find x round to the nearest tenth

jeka94

<h3>Answer: angle X = 70.5 degrees</h3>

=================================================

Work Shown:

Law of Cosines

c^2 = a^2 + b^2 - 2ab*cos(C)

22^2 = 20^2 + 18^2 - 2*20*18*cos(X)

484 = 724 - 720*cos(X)

484 + 720*cos(X) = 724

720*cos(X) = 724 - 484

720*cos(X) = 240

cos(X) = 240/720

cos(X) = 1/3

X = arccos(1/3)

X = 70.528779

X = 70.5

Make sure your calculator is in degree mode.

7 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

3 years ago

Andrew is showing his work in simplifying (−5.7 + 2.2) − 1.1.

Leviafan [203]

Answer: In step 1, Andrew used the commutative property

Explanation:
In step 1 he used commutative, which is a + b = b + a
(- 5.7 + 2.2) = 2.2 + (- 5.7)

Step 2, he used associative property, not the distributive.

Step 3, he just added5.7 + 1.1 = 6.8

Step 4, he distributed the negative sign:
- (6.8) = - 6.8

6 0

2 years ago

Read 2 more answers

Write an algebraic expression for this situation . Steve made a down payment of $300 on a new tv. He paid $37 per month for x mo

Temka [501]

300+37x
300+37(9)
300+333
633

5 0

3 years ago