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3 years ago

Graph f(x) = 22 +1 and g(x) = -x +4 on the same graph

Mathematics

1 answer:

Virty [35]3 years ago

7 0

Answer:

$-19 = x$

Step-by-step explanation:

22 + 1 = -x + 4

23 = -x + 4

- 4 - 4

________

19 = −x

$-19 = x$

I am joyous to assist you anytime.

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To calculate the cost of painting his silo, a farmer must find its height. The farmer uses a cardboard square to line up the top

tankabanditka [31]

Answer:

Height of the silo = 18 feet.

Step-by-step explanation:

From the figure attached BC is the length of the silo and the height of the farmer is 5 ft.

Farmer is standing at 8 ft distance from the silo.

From triangle AEC,

tan(∠CAE) = $\frac{CE}{AE}$

= $\frac{5}{8}$

m(∠CAE) = $tan^{-1}(\frac{5}{8})$

= 32°

m∠BAE = 90° - 32° = 58°

From the triangle ABE,

tan58° = $\frac{BE}{AE}$

BE = 8tan58°

BE = 12.8 ft

Total height of the silo = BE + EC

= 12.8 + 5

= 17.8

≈ 18 ft

Therefore, total height of the silo is 18 ft.

3 0

2 years ago

Match the vocabulary to the part of expression below

Fantom [35]

Answer:

Gas up in my room I don’t know how much you have to do that yet but

Step-by-step explanation:

Abc news I didn’t see you yesterday I didn’t answer you I didn’t answer your phone I didn’t answer your question I don’t understand why

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2 years ago

Henry bought a laptop for 4500 the cost of the laptop deprecate by 6% every year.If he decided to sell the laptop after 4 years

damaskus [11]

Answer:

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Step-by-step explanation:

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3 years ago

If two straight lines intersect and one angle is 30 degrees what are the other three angles

mel-nik [20]

Easey peasy
oposite angle is equal
other angles are supplementary (supplementary means add to 180, and a straight line=180 so if you look at 2 lines intersecting and think for a little bit, you wiill see why)

oposite angle=30
other angle is adds to 180
30+other=180
subtract 30
other=150
other is oposite other other
other=other other
150=other other

angles are
150,30,150

3 0

2 years ago

Read 2 more answers

Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.

professor190 [17]

Answer:

The solution to the differential equation is

$X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)$

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

$x''(t)+4x'(t)+6x(t)=1-e^t$

So we will get

$s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

$s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Simplifying

$s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}$

$s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Moving all terms that do not have X(s) to the other side

$s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5$

Factoring X(s) and moving the rest to the other side.

$X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5$

$X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}$

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

$\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}$

So if we multiply both sides by the entire denominator we get

$1=A(s^2+4s+6) + (Bs+C)s$

At this point we can find the value of A fast if we plug s = 0, so we get

$1=A(6)+0$

So the value of A is

$A = \cfrac 16$

We can replace that on the previous equation and multiply all terms by 6

$1=\cfrac 16(s^2+4s+6) + (Bs+C)s$

$6=s^2+4s+6 + 6Bs^2+6Cs$

We can simplify a bit

$-s^2-4s= 6Bs^2+6Cs$

And by comparing coefficients we can tell the values of B and C

$-1= 6B\\B=-1/6\\-4=6C\\C=-4/6$

So the separated fraction will be

$\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}$

We can repeat the process for the second fraction.

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}$

Multiplying by the entire denominator give us

$1=A(s^2+4s+6) + (Bs+C)(s-1)$

We can plug the value of s = 1 to find A fast.

$1=A(11) + 0$

So we get

$A = \cfrac1{11}$

We can replace that on the previous equation and multiply all terms by 11

$1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)$

$11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C$

Simplifying

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C$

And by comparing coefficients we can tell the values of B and C.

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}$

So the separated fraction will be

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}$

So far replacing both expanded fractions on the solution

$X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}$

We can combine the fractions with the same denominator

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}$

Simplifying give us

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}$

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

$s^2+4s+6 = s^2 +4s+4-4+6$

We are adding half of the middle term but squared, so the first 3 terms become the perfect square, that is

$=(s+2)^2+2$

So we get

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}$

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Lastly we can split the fraction one more

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

$\boxed{X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}$

6 0

3 years ago