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yawa3891 [41]
3 years ago
10

I need help please i would appreciate it​

Mathematics
2 answers:
Tju [1.3M]3 years ago
6 0

Answer:

5/1 × 1/3 = 5/3

Step-by-step explanation:

correct me if im mistaken,but im sure my answer is correct

Reptile [31]3 years ago
3 0
Answer: 5/3

i hope that’s correct :)
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Which of the following number lines represents the inequality below?
zvonat [6]

Answer:

B

Step-by-step explanation:

The circle has to be hallow in order for it to be greater than or less than. The arrow is going below 4 because the circle is on 4 and the arrow is pointing to the left side.

8 0
2 years ago
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A factory makes candles. Each candle is in the shape of a triangular prism, as shown below. If the factory made 35 such candles,
Llana [10]
Number of candles =(total volumes used by the company)/(volume of one candle)
volume of one candle =5cm*2cm*12cm=120cm^3
hence number of candles =5040/120
, ,=42candles
5 0
3 years ago
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
√72 what is the non-perfect simplefied square root radical
Lostsunrise [7]

For simplifying radicals, remember the product rule of radicals: √ab = √a x √b . This radical is simplified as such:

√72 = √9 x √8 = 3 x √4 x √2 = 3 x 2 x √2 = 6√2

In short, the simplified radical of √72 is 6√2.

5 0
3 years ago
What's the answer ?
koban [17]
A=4*<span>π*r^2
A=4*</span><span>π*2^2
=50.3</span>
7 0
4 years ago
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