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3 years ago

A lorry driver charges $1800 to carry 60 tonnes of weight. How much will he charge for carrying 80 tonnes at the same rate

Mathematics

2 answers:

V125BC [204]3 years ago

4 0

Answer:

$2400

Step-by-step explanation:

60 tonnes = $1800

80 tonnes = $?

= 1800 × 80/60

= $2400

ki77a [65]3 years ago

4 0

Step-by-step explanation:

$2400 to Carry 80tonnes of weight

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last month leonardo went to the gas station 5 times to buy gas for his car the numbers of gallons he purchased were 14.071, 14.3

zavuch27 [327]

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2 years ago

Suppose you discover that your growth fund, income fund, and money market fund return

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Answer:−3+3=−4

2+3−=15

4−3−=19

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Step 1: Pair the equations to eliminate y because the y terms are already additive inverses

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Step 2: Write the two new equations

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Step 3: Substitute the value for x and z

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15+2=11

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Step-by-step explanation:

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3 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,

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