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QveST [7]
3 years ago
5

Solve a + 14 < 20. solve 6 grade math​

Mathematics
2 answers:
AnnyKZ [126]3 years ago
7 0

Answer:

a<6

Step-by-step explanation:

You need to get "A" by its self so you subtract 14 from both sides and you get the answer "A" equals less than six.

gayaneshka [121]3 years ago
7 0

Answer:

 A. can equal any number below 6

Step-by-step explanation:

There is no way to tell exactly what number a is equal to with the information given, but we do know that since  a + 14 is less than 20, a can't equal 6 or any number above it.

Hope that makes sense! :)

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Mamont248 [21]
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73.50 x 3.50 = 21 hours
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In your own words, explain why there is no solution to the following equation:| x-6|= -8 Also, knowing this, how does it help to
solniwko [45]

Answer:

Step-by-step explanation:

So the two lines before and after the expression means absolute value, or modulus of, knowing this, it means that the answer must always yield positive. So if x-6 is positive, it will stay positive, if x-6 is negative, it will turn positive, therefore it can never yield a negative value.

Now im assuming the second question is meant to be absolute value of x-5 is less than 0, because it makes no sense otherwise.

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3 years ago
7717 as a percentage
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3 years ago
Write 55 as a product of its prime factors
Roman55 [17]

Answer:

55 = 5 × 11

Step-by-step explanation:

The only prime factors of 55 are 5 and 11, thus

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4 0
3 years ago
Read 2 more answers
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
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