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natima [27]
2 years ago
12

Given: g(a) = 3a +2 and f(a)= a2 - 1 Find: f(a-1)= g(a-1)

Mathematics
1 answer:
VARVARA [1.3K]2 years ago
3 0

Answer:

a=-2

Step-by-step explanation:

Though I am unsure if we were to find a but yeah, hope it's correct

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If an orthocenter lies inside of a triangle, then the triangle must be
Black_prince [1.1K]

Answer:

<h2><u><em>acute</em></u></h2>

Step-by-step explanation:

Remember :the orthocenter of  a triangle is the point of concurrency of the 3 altitudes.

\large \text{If an orthocenter lies inside of a triangle, then the triangle must be}\ :\\\\\large isosceles. \\\\\large obtuse. \\\\\large right. \\\\\large acute \checkmark

8 0
2 years ago
What percent of 143 is 95.1?
Aloiza [94]

Answer:

66.5%

Step-by-step explanation:

95.1/143*100=66.5%

4 0
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
Larry had 3⁄7 of a strawberry-rhubarb pie left over. He split the leftover pie evenly between his 3 children. What fraction of a
Damm [24]

Answer:

1/7

Step-by-step explanation:

The left over fraction is divided among the three children as follows

Given

Left over as 3/7

Number of children 3

Quantity each gets will be 3/7÷3

3/7÷3=3/7*1/3=3/21

By simplyfying the fractions, 3/21 when both the numerator and denominator are divided by 3 we obtain that 3/21=1/7

Therefore, each child geta 1/7

7 0
3 years ago
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r-ruslan [8.4K]
3.987 rounded to the nearest tenth = 4.0
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3 years ago
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