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3 years ago

PLEASE HELP, 50 POINTS QUESTION IS IN THE PICTURE

Mathematics

2 answers:

Sergio [31]3 years ago

6 0

Answer:

- 50/3

Step-by-step explanation:

artcher [175]3 years ago

4 0

Answer:

$-\frac{50}{3}$

Step-by-step explanation:

Find dy/dx of the ellipse $25x^{2} +y^{2}=109$ at (2,3)

Using Implicit Differentiation

Differentiate both sides of the equation with respect to x

$25x^{2} +y^{2}=109$

$50x +2y\frac{dy}{dx}=0$

isolate the term dy/dx

$2y\frac{dy}{dx}=-50x$

$\frac{dy}{dx}=-25\frac{x}{y}$

Evaluate at point (2,3)

For x=2, y=3

$\frac{dy}{dx}=-25(\frac{2}{3})$

$\frac{dy}{dx}=-\frac{50}{3}$

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2 years ago

Which is the simplified form of the expression?

Scorpion4ik [409]

Answer:

1 / q^30.

Step-by-step explanation:

[(p^2)(q^5)]^-4 * [(p^-4)(q^5)]^-2

Using the law (a^b)^c = a^bc :-

= p^-8 * q^-20 * p^8 * q^-10

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8 0

2 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

2 years ago

James knows that when he walks, he takes about 120

Tpy6a [65]

Answer:

speed of James is: 3.75 miles/hour.

Step-by-step explanation:

We need to find the speed of James in miles per hour(mi/hr)

we know that 1 mile=5280 feet

James takes 120 steps in 1 minute so he will cover 120×60 steps in 1 hour.

i.e. he takes 7200 steps in 1 hour.

also each step of James=2.75 feet

total distance covered by James in 1 hour=2.75×7200 feet=19800 feet

$19800feet=\frac{19800}{5280} miles=3.65miles$

hence speed of James= (distance covered by james in 1 hour)/1 hour

speed of James=3.75 miles/hour.

Thanks!!

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7 0

2 years ago

Read 2 more answers

Kyle spend 1/5 of his money on clothing. He had 120 remaining. What was the cost of the clothes Kyle bought

dolphi86 [110]

Answer:

30 is the cost of clothes Kyle bought.

Step-by-step explanation:

Given: Kyle spend 1/5 of his money on clothing.

Remaining money is 120.

Lets assume the total money Kyle had be "x".

∴ Money spent on clothes = $\frac{1}{5}\times x= \frac{x}{5}$

Now, putting up value in the equation to find the total amount of money.

We know, money remaining is 120

$x-\frac{x}{5} = 120$

Taking LCD subtract the fraction.

⇒ $\frac{(x\times 5)- x\times 1}{5} = 120$

⇒ $\frac{5x-x}{5} = 120$

⇒ $\frac{4x}{5} = 120$

Multiplying 5 both side

⇒ $4x= 120\times 5$

dividing 4 on both side.

∴ $x= \frac{120\times 5}{4}$

∴ x= 150

Kyle has total 150 money to spend.

Kyle spent $\frac{1}{5}$ of his total money on clothing.

⇒ cost of clothes kyle bought= $\frac{1}{5} \times 150= 30$

Hence, Kyle spent 30, which was the cost of clothes kyle bought.

6 0

2 years ago