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3 years ago

5

-2(x - 2) - 4x = 3(x + 1) - 9x

1 answer:

Sonbull [250]3 years ago

4 0

$\huge\boxed{\text{no solution}}$

Distribute the $-2$ and the $3$ . $-2x-2(-2)-4x=3x+3(1)-9x$

Multiply. $-2x+4-4x=3x+3-9x$

Combine the like terms. $-6x+4=-6x+3$

Add $6x$ on both sides of the equation. $4=3$

This statement is false, so this equation has $\boxed{\text{no solution}}$ .

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crimeas [40]

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3 years ago

The two parallel chords in circle C have lengths of 8 and 12. The distance (BD) between the chords is 3. Find the radius.

aivan3 [116]

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4 years ago

When Mrs. JHixson gave a test, the scores were normally distributed with a mean of 78 and a standard deviation of 6. This means

Sergio039 [100]

For normal distribution 95% lie in the range mean-2*standard deviation for lower bound and mean+2*standard deviation for upper bound.

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8 0

3 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

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

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

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

or

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

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.

7 0

3 years ago

MissTica

The answer is B 3x^4-13x^3-x^2-11x+6

7 0

3 years ago