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Novosadov [1.4K]
2 years ago
12

A linear function has a slope of Negative StartFraction 7 Over 9 EndFraction and a y-intercept of 3. How does this function comp

are to the linear function that is represented by the equation y + 11 = Negative StartFraction 7 Over 9 EndFraction (x minus 18)? a It has the same slope and the same y-intercept. b It has the same slope and a different y-intercept. c It has the same y-intercept and a different slope. d It has a different slope and a different y-intercept.
Mathematics
1 answer:
dem82 [27]2 years ago
7 0

It has the same slope and the same y-intercept, that is, <u>option a</u> is the right choice.

A linear function in the slope-intercept is written as y = mx + b, where m represents the slope of the line and b represents the y-intercept.

We are given two linear functions and are asked to compare them.

<u>First </u><u>linear function:</u>

Slope (m) = -7/9.

y-intercept (b) = 3.

Therefore, the linear function is: y = (-7/9)x + 3.

<u>Second </u><u>linear function</u>:

Function given: y + 11 = (-7/9)(x - 18)

or, y + 11 = (-7/9)x + (-7/9)(-18)

or, y = (-7/9)x + 14 - 11 = (-7/9)x + 3.

Therefore, slope (m) = (-7/9)

y-intercept (b) = 3.

Therefore, we can say that it has the same slope and the same y-intercept, that is, <u>option a</u> is the right choice.

Learn more about linear functions at

brainly.com/question/15602982

#SPJ10

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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0
3 years ago
I’m pretty confused on this question. Anyone know the answers?
Afina-wow [57]

Answer:

see explanation

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

• If m > 0, then the line slopes upwards from left to right ( increases )

• If m < 0, then the line slopes downwards from left to right ( decreases )

(a)

y + x = 3 ( subtract x from both sides )

y = - x + 3 ← in slope-intercept form

with m = - 1 ⇒ decreasing from left to right

(c)

y = 4x + 8 ← is in slope-intercept form

with m = 4 ⇒ increasing from left to right

the equation of a line in the form x = c represents a vertical line, where c is the value of the x-coordinates the line passes through

(b)

x - 5 = 0 ( add 5 to both sides )

x = 5 ← is a vertical line

the equation of a line in the form y = c represents a horizontal line, where c is the value of the y-coordinates the line passes through

(d)

\frac{y}{4} = - 6 ( multiply both sides by 4 )

y = - 24 ← is a horizontal line


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3 years ago
Nora is in the business of manufacturing phones. She must pay a daily fixed cost to rent the building and equipment, and also pa
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9514 1404 393

Answer:

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Step-by-step explanation:

We know the variable cost is 150 per phone, so the total cost will be ...

  c = 150p +b . . . . for some value b

We also know the value of c for 8 phones, so we can find b:

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Then the desired equation is ...

  c = 150p + 1000

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5x² + |x+1| > 0
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