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ss7ja [257]
3 years ago
14

What is the slope of the line that passes through the points (-2,8) and (-22,8)

Mathematics
2 answers:
Vanyuwa [196]3 years ago
6 0

Answer: m = 0

Step-by-step explanation:

Lana71 [14]3 years ago
4 0

Answer:

m=0

Step-by-step explanation:

The slope formula is:  m=y1=y2/x2=x1

where (x₁, y₁) and (x₂,y₂) are the points on the line.

The points given are (-2,8) and (-22,8). Therefore,

x1=-2

y1=8

x2=-22

y2=8

Substitute the values into the formula.

m=8-8/-22-2

Solve the numerator. Subtract 8 from 8.

m=0/-22+2

Zero divided by any number is zero, so we automatically know the slope is zero.

m=0

The slope of the line is 0

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Scorpion4ik [409]
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The sigma notation would be:
∑ (from n=1 to ∞) 4800 * (1/4)ⁿ , where n is the year. 
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This means the total number of flowers equals:
Year 1: 4800 * 1/4 = 1200 ]
+
Year 2: 4800 * (1/4)² = 300
+ 
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Year 4: 4800 * (1/4)⁴ = 18.75 = ~19 (we can't have a part of a flower)
+
Year 5: 4800 * (1/4)⁵ = 4.68 = ~ 5
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1200 + 300 + 75 + 19 + 5 + 1 = 1,600

8 0
3 years ago
F(x) = x2 what is g(x)?
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Answer:

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Step-by-step explanation:

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3 years ago
Violet made cookies for the bake sale. 7/10 of the cookies are chocolate chip. 3/4 of the remaining cookies are sugar. What frac
FromTheMoon [43]
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3 0
2 years ago
A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
2 years ago
Need help asap! will give the first right answer brainliest!
stiv31 [10]

Answer:

Step-by-step explanation:

12

4 0
2 years ago
Read 2 more answers
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