All categories

3 years ago

Can I get some help with this problem

Mathematics

1 answer:

Andrej [43]3 years ago

5 0

Answer:

I cant give you the answer but I do know that you need a compass to solve this. ( the first corner is 70degrees)

You might be interested in

I need help with this question I don't know if you can see it

Dovator [93]

$\begin{gathered} (-5,-6)\Rightarrow C \\ (1,-7)\Rightarrow K \\ (4,4)\Rightarrow A \\ (-6,-2)\Rightarrow F \\ \text{Star}\Rightarrow(-9,3) \\ Heart\Rightarrow(-5,-9) \\ \text{Lightning}\Rightarrow(3,-8) \\ Triangle\Rightarrow(1,-8) \end{gathered}$

5 0

1 year ago

Find the indefinite integral. (use c for the constant of integration.) sin3 5θ cos 5θ dθ

Elina [12.6K]

I'm guessing you mean

$\displaystyle\int\sin^35\theta\cos5\theta\,\mathrm d\theta$

Let $t=5\sin\theta$ . Then $\mathrm dt=5\cos5\theta\,\mathrm d\theta$ . So we have

$\displaystyle\int\sin^35\theta\cos5\theta\,\mathrm d\theta=\frac15\int\sin^35\theta(5\cos5\theta)\,\mathrm d\theta$

$=\displaystyle\frac15\int t^3\,\mathrm dt$

$=\dfrac{t^4}{20}+C$

$=\dfrac{\sin^4\5\theta}{20}+C$

7 0

3 years ago

In rectangle ABCD, AB=8 units and BC=15 units. Find BD

BabaBlast [244]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

45 points.!!!! <br> Does a bias study mean the conclusions reached by the study are false?

miv72 [106K]

It doesnt mean they are false it just means the person giving the study is going to favor something over something else in the study. Hes basically picking favorites.

3 0

3 years ago

Read 2 more answers

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago

Can I get some help with this problem ​

Can I get some help with this problem