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ss7ja [257]
3 years ago
14

Help meeeeee, i don’t understand

Mathematics
1 answer:
Pavel [41]3 years ago
7 0

Answer:

To graph the equation, you need to find the slope and y-intercept.

y = mx + b

⬑<em>[slope-intercept formula; m is the slope, and b is the y-intercept]</em>

y = 4x + −1

⬑<em>[given equation before being simplified]</em>

y = 4x − 1

⬑<em>[given equation]</em>

║m = 4

║b = (0, −1)

Proof:

m = \frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1}

x₁ = 0

x₂ = 1

y₁ = −1

y₂ = 3

m = \frac{3 - -1}{1 - 0}

║m = 4

║b = (0, −1)

<em>[the value for x₁ and y₁ equals the </em><em>y-intercept: (0, </em>−<em>1)</em><em>]</em>

Graph:

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1.625

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How many different anagrams can you make for the word MATHEMATICS?
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Answer:

4989600

Step-by-step explanation:

The given word is MATHEMATICS

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the given word is arranged in 11! ways

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The integer between 42 and 140
Montano1993 [528]
It has 97 amounts.
43,44,45,46,47,48,49,50,
51,52,53,54,55,56,57,58,59,60
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91,92,93,94,95,96,97,98,99,100
101,102,103,104,105,106,107,108,109,110
111,112,113,114,115,116,117,118,119,120
121,122,123,124,125,126,127,128,129,
130
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6 0
3 years ago
HELP PLSPLSPLS
ddd [48]

all "atomic" or constituent statements are true

at least one "atomic" or constituent statement is true

3 0
3 years ago
A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min
Dmitry [639]

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

6 0
4 years ago
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