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3 years ago

Please help me i beg you!!! 30 points!!!

Mathematics

2 answers:

Westkost [7]3 years ago

6 0

The third is because the prove is x = fifteen and if a number is next to a letter, it means multiply, sorry I could only do one.

mash [69]3 years ago

3 0

I hope this helps you

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SOMEONE PLEASE HELP!

Andrews [41]

Answer:

17.2°

Explanations

To find angle A, use the cosine rule.

a^2 = b^2 + c^2 - 2 × a × c cos A

4^2 = 7^2 + 10^2 - 2 × 7 × 10 cos A

16 = 49+ 100 - 140 × cosA

16 = 149 - 140cosA

16- 149 = - 140cosA

-133 = - 140cosA

cosA = 133/140

cosA = 0.95

A = 17.2°

7 0

2 years ago

Solve for z -p(d+z) = -2z+59

Aleksandr-060686 [28]

Here you go. Let me know if you have questions

3 0

3 years ago

angle and angle b are complentary angles. if angle a = (2x-5) and m angle b = (5x+4) then find the measure of angle a

iris [78.8K]

complementary angles = 90

2x-5 + 5x+4 = 90

combine like terms

7x-1 =90

7x = 91

divide by 7

x =13

angle a

2x-5

2*13-5

26-5

Angle a: 21

6 0

3 years ago

(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

3 years ago

Which equation could be used to solve the following system?

nikdorinn [45]

we have

$x+y=11$

$x=11-y$ --------> equation A

$4x^{2} -3y^{2}=8$ ----> equation B

Substitute equation A in equation B

$4(11-y)^{2} -3y^{2}=8$

therefore

<u>the answer is the option B</u>

$4(11-y)^{2} -3y^{2}=8$

8 0

3 years ago

Read 2 more answers