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2 years ago

Solve for x: 5x > -20

Mathematics

1 answer:

konstantin123 [22]2 years ago

5 0

The answer is x > -4

To get this answer, you divide both sides by 5. This is to undo the multiplication that happens to the 'x' in '5x'. Note that 5x means 5*x or 5 times x, where x is just a placeholder for some unknown number.

Saying x > -4 means you can pick any number larger than -4.

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1 pound is closest in weight to which one of the following Metric units of weight? 450 grams 4,500 centigrams 0.045 kilograms 45

ch4aika [34]

Answer:

0.045 kilograms

Step-by-step explanation:

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2 years ago

-3(x - 2) - (2x - 3) = 5 - (4x - 2)

STatiana [176]

Answer:

x=2

Step-by-step explanation:

-3*(x-2)-(2*x-3)-(5-(4*x-2))=0

((0 - 3 • (x - 2)) - (2x - 3)) - (7 - 4x) = 0

2 - x = 0

-x+2 = 0

Subtract 2 from both sides of the equation :

-x = -2

Multiply both sides of the equation by (-1) : x = 2

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2 years ago

Read 2 more answers

If the pattern in the table continues when will sales for DVD players equal sales for mobile phones?

omeli [17]

Answer:

<em>never </em>because they are continually increasing at the same rate without multiplying and dividing

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2 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$