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lianna [129]
2 years ago
15

Write 22 as a product of primes

Mathematics
2 answers:
ANEK [815]2 years ago
7 0

Answer:

2•11

Step-by-step explanation:

2 and 11 are both prime numbers and can be multiplied together to get 22

Inga [223]2 years ago
3 0

Answer:

2 x 11

Step-by-step explanation:

Hope that helps!!!!

<em>-scsb17hm</em>

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How do I find the value of x?
svetoff [14.1K]

Answer:

             x = 2

Step-by-step explanation:

If ΔABC~ΔAED then:

                                 \dfrac{AC}{AD}=\dfrac{BC}{ED}\\\\\\\dfrac{x+4}{4}=\dfrac{3}{2}\\\\\dfrac{x+4}{4}=\dfrac64\\\\x+4=6\\\\x=2

4 0
2 years ago
Sam purchased a used car for $15000. He paid 6.3% sales tax. How much tax did he pay?
atroni [7]

Answer:

$945

Step-by-step explanation:

($15000)(.063) = $945

*note: divide by 100 to turn percent into decimal

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3 years ago
Determine whether or not the given procedure results in a binomial distribution. If not, identify which condition is not met. Su
harkovskaia [24]

Answer:

The answer is "No, There are more than two possible outcomes on each trial of the experiment ".

Step-by-step explanation:

When various ice cream products are known. This might surpass 2 brands or more. Thus the number of different results varies considerably.

BINOMIAL DISTRIBUTION:

An investigation with a set set of individual tests, each only with two possible results.

Four conditions are met by the binomial experiment

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7 0
2 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
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Answer:

to be honost i am in 8th grade but i realy like you and your the first person i like in a lot of time

Step-by-step explanation:


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