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NeX [460]
3 years ago
9

Can you solve this please ;)

Mathematics
2 answers:
Soloha48 [4]3 years ago
7 0
It is c for you answer
AleksAgata [21]3 years ago
3 0
C) X is greater than or equal to 0

(Domain is all the x-values the line goes through)

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Find the area of the shape shown<br> below.
steposvetlana [31]

Answer:

The area of the figure is 28 units².

Step-by-step explanation:

Find the area of the full rectangle, ignoring the triangle.

a = lw \\ a = 4 \times 8 = 32

Then, find the area of the triangle:

a =  \frac{bh}{2}  \\ a =  \frac{2 \times 4}{2}  \\ a = 4

Finally, subtract:

32 - 4 = 28

7 0
3 years ago
The question is in the picturere
Xelga [282]

Answer:

12

Step-by-step explanation:

you subtract the ordered pairs on the sides. then the ones on the top and bottom. then you add all of the sides

4 0
3 years ago
Explain how you know if a system of equations has one solution, no solutions, or infinitely many solutions. Write your answers (
Nataliya [291]

Answer:

Written Below.

Step-by-step explanation:

A system of linear equations has one solution when the graphs intersect at a point. A system of linear equations has no solution when the graphs are parallel. A system of linear equations has infinite solutions when the graphs are the exact same line.

(This uses graphing.)

Sidenote: I hope this helps! :)

7 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
Goodmorning hope you have a great day :)
zalisa [80]

Answer:

hello good morning.have a nice day

6 0
2 years ago
Read 2 more answers
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