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Degger [83]
3 years ago
14

What is the slope of the line passing through the points (1, −5)(1, −5) and (4, 1)(4, 1)?

Mathematics
1 answer:
galina1969 [7]3 years ago
4 0
The answer to this is 2.
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19-(3-5b)<br>....,.......​
Serjik [45]

Answer:

Step-by-step explanation:

19(-3+5b)

-57+95b

6 0
4 years ago
What is the eighth term of the arithmetic sequence? An=<br> 10n – 14?
Alja [10]

Answer:

A8 = 66 [8th term]

Step-by-step explanation:

Given the explicit arithmetic sequence An = -14 + 10n → -4 + 10(n-1)

To find the 8th term. An nth term is where n is n in the sequence. So substitute 8 for n, and simplify.

So A8 = -4 + 10(8-1) → -4 + 10(7) → -4 + 70 → 70 - 4 → 66.

3 0
3 years ago
22. Solve the system
DaniilM [7]
Equation 1)  x + 6y = 2
Equation 2)  5x + 4y = 36

Multiply all of equation 1 by 5.

1)  5(x + 6y = 2)

Simplify.

1)  5x + 30y = 10
2)  5x + 4y = 36

Subtract equations from one another.

26y = -26

Divide both sides by 26.

y = -1

Plug in -1 for y in the first equation.

x + 6y = 2

x + 6(-1) = 2

Simplify.

x - 6 = 2

Add 6 to both sides.

x = 8

D : (8, -1)

~Hope I helped!~
8 0
3 years ago
Which of the following functions represents a cell phone bill if there is a monthly charge of 20 dollars plus 6 cents for each m
pychu [463]

Answer:

f(x)=20+0.06x

Step-by-step explanation:

3 0
3 years ago
A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th
Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let S(t) be the amount of sugar in the tank at time t. Then S(0)=25.

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}

Separate variables, integrate, and solve for <em>S</em>.

\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt

\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt

-100\ln\left|P-\dfrac S{100}\right|=t+C

\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t

P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}

\dfrac S{100}=P-Ce^{-100t}

S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}

Use the initial value to solve for <em>C</em> :

S(0)=25\implies 25=100P-C\implies C=100P-25

\implies S(t)=100P-(100P-25)e^{-100t}

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}

5 0
3 years ago
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