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Irina18 [472]
3 years ago
15

Answer this question please! 25 points and brainliest!

Mathematics
2 answers:
Eva8 [605]3 years ago
8 0

Answer:

c    x< 9

Step-by-step explanation:

A number less than 9

x must be less than 9

No equals sign because it is less than

The inequality faces the larger number

x < 9

Amiraneli [1.4K]3 years ago
4 0

Answer: c


Step-by-step explanation:

the number with one dot < on the left side means its the smaller number

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Which of these numbers are less than 8.1 × 10^-8?
lina2011 [118]
The answer is 3.4×10^-10
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3 years ago
For example( and this is straight from the paper), "15 people in 3 equal rows."
densk [106]
At a race, there are 3 rows with 15 people in each row. How many total people are there?

Numerical expression- 15 times 3

(Idk if this helped very much but it's worth a shot)
6 0
3 years ago
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
Find the surface area of the following cylinders. Round to the nearest hundredth.
Aleks [24]

Answer:

surface area of cylinder= 2πrh + 2πr²

2×22/7×1×8+2×22/7×1×1

<u>352</u><u>.</u> + <u>4</u><u>4</u>

7. 7

= 50.3 + 6.25

= 56.55cm²

5 0
3 years ago
How do i solve -8=-(x+4)
igomit [66]
Distribute the implied one, making it -8=-x-4 then add four to each side making it -4=-x divide by -1 and the answer will be 4=x
4 0
3 years ago
Read 2 more answers
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