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3 years ago

Pls help due in 15 mins!!

Mathematics

2 answers:

WITCHER [35]3 years ago

8 0

Answer:

to late

Step-by-step explanation:

ella [17]3 years ago

7 0

your first constant of propotionality is 10

The second constant of propotionality 5

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What is the value of x? A. 112 B. 68 C. 62 D. 56

ollegr [7]

Answer:

B 68

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3 years ago

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Ghella [55]

Answer:

26m

Step-by-step explanation:

8m x 2 = 16

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3 years ago

The vertices of △JKL are J(−1, 8), K(3, 5), and L(6, −9). What are the vertices of T<4, −2>(JKL)?

mestny [16]

J'=(3, 6); K'=(7, 3); L'=(10, -11)

This is a translation vector that shifts the figure 4 right and 2 down. We add 4 to each x-coordinate and subtract 2 from each y-coordinate.
J'=(-1+4, 8-2) = (3, 6)
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4 years ago

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and

allsm [11]

(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$ . The tank starts with only pure water, so $\boxed{A(0)=0}$ .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(A(t)/1080 kg/L) * (7 L/min) = 7A(t)/1080 kg/min

so that the net rate of change of $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$ :

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$ , we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.

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3 years ago

1/6(x + 6) = 11 What is x

frozen [14]

Answer:

x = 60

Step-by-step explanation:

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