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ICE Princess25 [194]
3 years ago
14

Please help I have 15 minutes!!!!!

Mathematics
2 answers:
Ber [7]3 years ago
7 0

Answer:

false

Step-by-step explanation:

Plug in the numbers given by the then statements. Plugging in -2 for x gives us -3 >= 3 which is false.

kumpel [21]3 years ago
7 0

Answer:

false

Step-by-step explanation:

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1+1<br> 1+2<br> 1+3<br><br><br><br> no one ever talks to me everyone just steals my points
marishachu [46]

Answer:

9

Step-by-step explanation:

8 0
3 years ago
98 POINTS!!!<br><br>Write a polynomial function in standard form given the zeros<br><br><br>±√5 , 7
Neporo4naja [7]

Answer:

Step-by-step explanation:

Knowing the three zeros of the equation, we can set it up as follows:

(x - \sqrt{5})(x + \sqrt{5})(x - 7)

Multipying the first two expressions together gives us the following:

(x^{2} - 5)(x - 7)

Multiplying the two expressions together gives us the following:

x^{3} - 7x^{2} - 5x + 35

3 0
3 years ago
Read 2 more answers
Write an equation for three-fourth of m is 20​
alisha [4.7K]

Answer:

Since, three fourth of \(t\) is \(15\). Hence, the required equation is \(\frac{3t}{4}=15\)

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Calc 3 iiiiiiiiiiiiiiiiiiiiiiiiiiii
Lilit [14]

Take the Laplace transform of both sides:

L[y'' - 4y' + 8y] = L[δ(t - 1)]

I'll denote the Laplace transform of y = y(t) by Y = Y(s). Solve for Y :

(s²Y - s y(0) - y'(0)) - 4 (sY - y(0)) + 8Y = exp(-s) L[δ(t)]

s²Y - 4sY + 8Y = exp(-s)

(s² - 4s + 8) Y = exp(-s)

Y = exp(-s) / (s² - 4s + 8)

and complete the square in the denominator,

Y = exp(-s) / ((s - 2)^2 + 4)

Recall that

L⁻¹[F(s - c)] = exp(ct) f(t)

In order to apply this property, we multiply Y by exp(2)/exp(2), so that

Y = exp(-2) • exp(-s) exp(2) / ((s - 2)² + 4)

Y = exp(-2) • exp(-s + 2) / ((s - 2)² + 4)

Y = exp(-2) • exp(-(s - 2)) / ((s - 2)² + 4)

Then taking the inverse transform, we have

L⁻¹[Y] = exp(-2) L⁻¹[exp(-(s - 2)) / ((s - 2)² + 4)]

L⁻¹[Y] = exp(-2) exp(2t) L⁻¹[exp(-s) / (s² + 4)]

L⁻¹[Y] = exp(2t - 2) L⁻¹[exp(-s) / (s² + 4)]

Next, we recall another property,

L⁻¹[exp(-cs) F(s)] = u(t - c) f(t - c)

where F is the Laplace transform of f, and u(t) is the unit step function

u(t) = \begin{cases}1 & \text{if }t \ge 0 \\ 0 & \text{if }t < 0\end{cases}

To apply this property, we first identify c = 1 and F(s) = 1/(s² + 4), whose inverse transform is

L⁻¹[F(s)] = 1/2 L⁻¹[2/(s² + 2²)] = 1/2 sin(2t)

Then we find

L⁻¹[Y] = exp(2t - 2) u(t - 1) • 1/2 sin(2 (t - 1))

and so we end up with

y = 1/2 exp(2t - 2) u(t - 1) sin(2t - 2)

7 0
2 years ago
Pq lies on plane PQR. Point S lies on plane PQR. qs does not lie on plane PQR. Which two statements contradict each other?
Vinil7 [7]

Answer c: 2 and 3


Of course Q is on plane PQR. If S is as well, then QS is as well.



8 0
3 years ago
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