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sveticcg [70]
3 years ago
14

A number has 6 hundreds, 7 tens, and 34 ones. What is the number?

Mathematics
2 answers:
s2008m [1.1K]3 years ago
8 0
600+ 70 + 34
600 + 104
704
coldgirl [10]3 years ago
7 0
Well 600 so thats there, now we have to add 70 and 34 which makes 104, so since "1" is the hundred we have to add 104 and 600 which gives you 704.
Answer: 704
 Hope i helped! :D
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-38/-8 plz help me plz​
Bad White [126]

Answer:

(-38) / (-8)

equals 4.75

a negative divided by a negative is always a positive

Step-by-step explanation:

5 0
3 years ago
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What is the equation in point slope form of the line that passes through the point (2,6) and has slope of 5?
sleet_krkn [62]
Firstly, you can use the slope and the first point to find a second point:
2 + 1 = x2 and 6 + 5 = y2 because the slope is 5/1. 
Next you can write the equation in point-slope form (remember point-slope form is y - y1 = m(x - x1):
y - 11 = 5(x - 3) 
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3 years ago
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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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poizon [28]
3 over 5 x minus 15 = 6 over 5 x plus 12

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8 0
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marusya05 [52]

Answer:

No

Step-by-step explanation:

Do they have common factors?

No which is why they are not able to be simplified further

so no

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