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3 years ago

What is the constant proportionality of y=32x-4

Mathematics

2 answers:

nalin [4]3 years ago

5 0

32 because it does not have a variable/letter. -4 the variable is x. and your saying y=32.

alexira [117]3 years ago

4 0

32 is the constant because it doesn’t have a letter

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What is the equation of the line that passes through (-9,12) and is perpendicular to the line whose equation is y=1/3x 6?

tigry1 [53]

Y = 1/3x ? 6....slope here is 1/3. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So our perpendicular line will have a slope of -3 (see how I flipped 1/3 and changed the sign)

y = mx + b
slope(m) = -3
(-9,12)...x = -9 and y = 12
now we sub and find b, the y int
12 = -3(-9) + b
12 = 27 + b
12 - 27 = b
-15 = b

so ur perpendicular equation is : y = -3x - 15

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3 years ago

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Neko [114]

Answer:

- 2 ≤ n < 8

Step-by-step explanation:

The closed circle at - 2 indicates that n can equal - 2

The open circle at 8 indicates that n cannot equal 8

Otherwise n can be all value between - 2 and 8 including - 2, thus

- 2 ≤ n < 8

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3 years ago

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Leno4ka [110]

Answer:

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Step-by-step explanation:

x equals to -3 is the standard form

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3 years ago

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Tom jogged 3/5 mile on Monday and 2/6 mile on Tuesday. how much farther did Tom jog on Monday than on tuesday

pashok25 [27]

3/5= 18/30, 2/6= 10/30, so Tom jogged 8/30 more of a mile on Monday then tuesday

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3 years ago

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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,

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

At t = 0

$C(0) = 8(e^{(0)}-e^{(0)}) = 0$

At t = 2

$C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185$

At t = 12

$C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059$

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

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2 years ago