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3 years ago

Can you find the slope of the line

Mathematics

1 answer:

Tatiana [17]3 years ago

5 0

Answer:

Slope is -3

Step-by-step explanation:

Using the formula (y2 - y1)/(x2 - x1), we can find that the ratio is -6/2. When simplified, it comes out to -3

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A scuba diver wants to descend at an average rate of at least 7 feet per minute for 5 minutes. The table shows how many feet the

gladu [14]

The scuba diver must descend 3 feet in the fifth minute to meet the goal.

Why?

To find how many feet does the scuba diver need to descend to reach the goal of 35 feet in 5 minutes, we need to calculate how many feet he did descend on the first 4 minutes.

So, calculating we have:

$FirstMinute=9ft\\SecondMinute=11ft\\ThirdMinute=5ft\\FourthMinute=7ft\\Total=9ft+11ft+5ft+7ft=32ft\\\\Goal=35ft\\\\Needed=Goal-Total(4Minutes)\\\\Needed=35ft-32ft=3ft$

Hence, we can see that at the 4 minute, he has only descended 32ft. It means that he need to descend 3 feet more in the fifth minute.

Have a nice day!

7 0

4 years ago

2.3×10⁹=(1×10³)(2.3×10^n)

kotegsom [21]

Answer:

n = 6

Step-by-step explanation:

2.3 × $10^{9}$ = (1 × 10³)(2.3 × $10^{n}$ )

(1 × 10³)(2.3 × $10^{n}$ ) = (1)(2.3) × (10³)( $10^{n}$ ) = 2.3 × $10^{3+n}$

2.3 × $10^{3+n}$ = 2.3 × $10^{9}$ ⇒ 3 + n = 9 ⇒ n = 6

5 0

3 years ago

Examine the function f(x)=x+(3/x). Find the point on the curve at which the tangent lines pass through the point (1, 1).

vazorg [7]

Base on the function that you give and the data that are given. The point on the curve at which the tangent lines pass through the point (1,1). Base on my calculation and through my analyzations i came up with an answer of -2x+3 = x+3/x

6 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

3 years ago

Leno4ka [110]

Answer:

6°F

58°F

Step-by-step explanation:

From the data Given :

Temperature rise in Buffalo by noon time = 14°F

The initial temperature at Buffalo = - 8°F

Rise in temperature = 14°F

Therefore, temperature at noon = (-8 + 14)°F = 6°F

B.)

Temperature at 6am in St. Louis = 32°F

Temperature at 6am in Juneau = - 26°F

Temperature difference :

(32 - (-26))°F

32 + 26

= 58°F

Temperature in St. Louis is 58°F higher

8 0

3 years ago