<img src="https://tex.z-dn.net/?f=-2cos%5E%7B2%7D%20x-6sinx-1%3D0" id="TexFormula1" title="-2cos^{2} x-6sinx-1=0" alt="-2cos^{2}

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musickatia [10]

3 years ago

x-6sinx-1=0" align="absmiddle" class="latex-formula">
Please help!!
Thank you!

Mathematics

1 answer:

seropon [69]3 years ago

6 0

<h3> Answer;</h3>

<h3> Domain x ER</h3>

<h3> Vertical intercept (0,--3</h3>

<h3> I solved it by graph and trust my</h3><h3>answer </h3>

like my answer and brainliest

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The Price family and the Jenkins family each used their sprinklers last summer. The water output rate for the Price family's spr

PIT_PIT [208]

Set up two equations:

1st: P (for Price family) + J (for Jenkins family) = 45 hours

2nd: 20P + 35J = 1200 liters

Rewrite the first equation as P = 45 – J

Replace P in the second equation:

20(45-J) + 35J = 1200

Simplify:

900 -20J + 35J = 1200

900 +15J = 1200

Subtract 900 from both sides:

15J = 300

Divide both sides by 15:

J = 300/15

J = 20

Jenkins used theirs for 20 hours

Price used theirs for 45-20 = 25 hours.

7 0

3 years ago

Find the solution of the given initial value problem: y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

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

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago

A laundry basket contains 10 socks, of which 2 are red. What is the probability that a randomly selected sock will be red? Write

algol13

Answer:

you get a pair of red socks. There are 10 socks in the drawer, and 2 are red. Therefore the probability that the first one will be red is 2/10 = 1/5.

Step-by-step explanation:

3 0

3 years ago

-2(x+2)=-12 .............

Dima020 [189]

Answer:

x=4

Step-by-step explanation:

-2(x+2)=-12

x+2=-12/-2

x+2=6

x=6-2

x=4

6 0

3 years ago

EASY 12 POINTS ❗️❗️

Taya2010 [7]

Answer:

A Water freezes in the cracks.

Step-by-step explanation:

Brainiest plz hope that help??

7 0

3 years ago