All categories

2 years ago

What is the measure of angle 15

Mathematics

2 answers:

Mademuasel [1]2 years ago

7 0

Answer:

Step-by-step explanation:

m<15 = m<1 because of how the transversals are placed. so m<1 = 90º - 65º = 25º

that means that m<15=25º

Mars2501 [29]2 years ago

3 0

Answer:

<15=25º

Step-by-step explanation:

big brain

You might be interested in

Which system of equations has no solutions? y -

Alona [7]

The second one on the top right

7 0

2 years ago

Complete the Table.

JulsSmile [24]

Interest rate = 5.25%

<u>Step-by-step explanation:</u>

Simple Interest = Pnr = 4252.50

n = time = 54 months = 4.5 years

r = rate of interest = ?

P = 18,000

r = SI/Pn = 4252.50 / 18000× 4.5

= 0.0525 = 5.25%

7 0

3 years ago

Read 2 more answers

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%20%2B%202x%7D%7B6x%20%5C%3A%7D%20%20%3D%20%20%5Cfrac%7B12%7D%7B5x%7D%20%20%2B%2

Reika [66]

Answer: x= 26/3

Step-by-step explanation: Solve the rational equation by combining expressions and isolating the variable x.

Let's solve your equation!

4 + 2x /6x = 12/5x

2x + 4/6x= 12/5x + 2/15

Multiply all terms by x and cancel:

2x + 4/6= 12/5
1/3x + 2/3 = 2/15x = 2/15x + 12/5 - 2/15x (Subtract 2/15x from both sides)
1/5x + 2/3 = 12/5
1/5x + 2/3 -2/3 = 12/5- 2/3 (subtract 2/3 from both sides)
1/5x = 12/5
Multiply both sides by 5
Check all your answers,
x= 26/3

8 0

3 years ago

Read 2 more answers

Question on laplace transform... number 7

irinina [24]

Recall a few known results involving the Laplace transform. Given a function $f(t)$ , if the transform exists, then denote it by $F(s)$ . We have

$\mathcal L_s\left\{e^{ct}f(t)\right\}=\mathcal L_{s-c}\left\{f(t)\right\}=F(s-c)$

$\mathcal L_s\left\{\displaystyle\int_0^t f(u)\,\mathrm du\right\}=\dfrac{F(s)}s$

$\mathcal L_s\left\{f'(t)\right\}=sF(s)-f(0)$

Let's put all this together by taking the transform of both sides of the ODE:

$y'(t)+2e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du=e^{-t}\sin t$

$\implies \bigg(sY(s)-y(0)\bigg)+2\mathcal L_s\left\{e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}=\dfrac1{(s+1)^2+1}$

Here we use the third fact and immediately compute the transform of the right hand side (I'll leave that up to you).

Now we invoke the first listed fact:

$\mathcal L_s\left\{e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}=\mathcal L_{s+2}\left\{\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}$

Let $g(u)=e^{2u}y(u)$ . From the second fact, we get

$\mathcal L_s\left\{\displaystyle\int_0^tg(u)\,\mathrm du\right\}=\dfrac{G(s)}s\implies\mathcal L_{s+2}\left\{\displaystyle\int_0^tg(u)\,\mathrm du\right\}=\dfrac{G(s+2)}{s+2}$

From the first fact, we get

$G(s)=\mathcal L_s\left\{e^{2u}y(u)\right\}=Y(s-2)$

so we're left with

$\dfrac{G(s+2)}{s+2}=\dfrac{Y((s+2)-2)}{s+2}=\dfrac{Y(s)}{s+2}$

To summarize, taking the Laplace transform of both sides of the ODE yields

$sY(s)+\dfrac{2Y(s)}{s+2}=\dfrac1{(s+1)^2+1}$

Isolating $Y(s)$ gives

$Y(s)=\dfrac1{\left(s+\frac2{s+2}\right)\left((s+1)^2+1\right)}$
$Y(s)=\dfrac{s+2}{\left((s+1)^2+1\right)^2}$

All that's left is to take the inverse transform. I'll leave that to you as well. You should end up with something resembling

$y(t)=\dfrac12(t\cos t-(t+1)\sin t)(\sinh t-\cosh t)$

4 0

3 years ago