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2 years ago

What is the slope below

Mathematics

2 answers:

raketka [301]2 years ago

4 0

Answer:

-1

Step-by-step explanation:

We can find the slope by change in y of change in x

change in y

---------------------

change in x

The y changes -3 ( goes down 3)

the x changes +3 ( to the right 3)

-3

----

This simplifies to -1

Butoxors [25]2 years ago

3 0

Answer:

the slope of the line is -1.

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The law of Cosines Is......

iogann1982 [59]

Answer: $\Large\boxed{C.~21}$

Step-by-step explanation:

The goal of the question

Find the value of 2abcosC

Given information

a = 4

b = 3

c = 2

Given formula (Law of Cosine)

$c^2=a^2+b^2-2abcosC$

Substitute values into the formula (Leave 2abcosC as it is)

$(2)^2=(4)^2+(3)^2-2abcosC$

Simplify the exponents

$4=16+9-2abcosC$

Simplify by addition

$4=25-2abcosC$

Add 2abcosC on both sides

$4+2abcosC=25-2abcosC+2abcosC$

$4+2abcosC=25$

Subtract 4 on both sides

$4 + 2abcosC-4=25-4$

$\Large\boxed{2abcosC=21}$

Hope this helps!! :)

Please let me know if you have any questions

5 0

1 year ago

Read 2 more answers

You are putting a rectangular hot tub on your back porch. It is 6 feet long, 5 feet wide, and 3 feet deep. How much will it cost

igor_vitrenko [27]

Answer:

$Cost=\$3.66234\approx \$3.67$

Step-by-step explanation:

$Height\ of\ tub=3\ feet\\\\Length\ of\ tub=6\ feet\\\\width\ of\ tub=5\ feet\\\\Volume=Length\times Width\times Height\\\\Volume=6\times 5\times 3=90\ feet^3$

$1\ feet^3=7.48052\ gallons\\\\90\ feet^3=90\times 7.48052=673.2468\ gallons$

$cost\ of\ 1000\ gallons=\$5\\\\cost\ of\ 1\ gallons=\$ \frac{5}{1000}\\\\cost\ of\ 673.2468\ gallons=\$ \frac{5}{1000}\times 673.2468=\$3.366234$

5 0

2 years ago

34. The product of two algebraic

S_A_V [24]

Answer:

3a²b

Step-by-step explanation:

The product of two algebraic

terms is 6a3b2. If one of the terms is

2ab, find the other term.

Let us represent

First term = a

Other term = b

a × b = 6a³b²

a = 2ab

b = ?

b = 6a³b²/a

b = 6a³b²/2ab

b = 6a³b²/2a¹b¹

b = (6 ÷ 2) × a^3 - 1 × b ^2 - 1

b = 3a²b

Therefore, the other term = 3a²b

7 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago

Yeah so help please.

irina [24]

Answer:

could you possibly zoom in

Step-by-step explanation:

7 0

2 years ago