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2 years ago

Please help!!!!!! Explain the answer

Mathematics

1 answer:

Dennis_Churaev [7]2 years ago

7 0

Answer:

I think they are.

Step-by-step explanation:

Because they are both 39 and 34 degrees so if x and y are the same number they are similar.

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A=1/2 (2x + 4) (6x + 4) solve using FOIL

boyakko [2]

A = 1/2 (12x² + 8x + 24x + 16)
A = 1/2 (12x² + 32x + 16)
A = 6x² + 16x + 8

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Answer: 6x² + 16x + 8
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7 0

2 years ago

Evaluate the expression when x = 2, y = 5, and z = 4.

REY [17]

Answer:

9 is the correct answer

(2×2+1)=5÷5×9=9

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2 years ago

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The temperature falls from 3 degrees Celsius to -4 degrees celcius what is the the difference in these temperatures

STatiana [176]

Answer:

Step-by-step explanation:

3 - -4=7

6 0

2 years ago

URGENT! 40 POINTS! PLEASE HELP!

umka2103 [35]

Answer:

(a) $8 x(5 x-6)=40 x^{2}-48 x$

Step-by-step explanation:

(a) To find verify the answer we need to multiplying the equation $8 x(5 x-6)$

$8 x(5 x-6)=40 x^{2}-48 x$

Thus, this statement is true.

(b) To find verify the answer we need to multiplying the equation $-4 x\left(2 x^{2}+1\right)$

$-4 x\left(2 x^{2}+1\right)=-8 x^{3}-4 x$

Hence, the given statement is false.

$\begin{aligned}(x-3)(5 x-6) &=5 x^{2}-6 x-15 x+18 \\&=5 x^{2}-21 x+18\end{aligned}$

Hence, the given statement is true.

(d) To find verify the answer we need to multiplying the equation $(2 x+3)\left(x^{2}+3 x-5\right)$

$\begin{aligned}=(2 x+3)\left(x^{2}+3 x-5\right) &=2 x^{3}+6 x^{2}-10 x+3 x^{2}+9 x-15 \\&=2 x^{3}+9 x^{2}-x-15\end{aligned}$

Hence, the given statement is false.

4 0

2 years ago

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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

2 years ago