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3 years ago

Don't understand number 1

Mathematics

1 answer:

Alexxx [7]3 years ago

7 0

It will be 20 cm I believe

Because AB is half of AC

if AB is 10 cm then the other half would be 10 cm

So just add those and you get 20cm

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PLEASE HURRY!! simplify completely show all work for full credit

Irina18 [472]

Answer:

$4x^{3}$

Step-by-step explanation:

Given:

$(\frac{192x^{12}}{3x^{3}} )^{\frac{1}{3}}$

Apply exponent rule of distribution:

$\frac{(192x^{12})^{\frac{1}{3}}}{(3x^{3})^{\frac{1}{3}}}$

Simplify the numerator:

$\\\\\frac{4 * 3^{\frac{1}{3}}x^{4}}{(3x^{3})^{\frac{1}{3}}}$

Simplify the denominator:

$\\\frac{4 * 3^{\frac{1}{3}}x^{4}}{3^\frac{1}{3}x}}$

Simplify:

$4x^{3}$

-> To explain this party since it is a bigger jump, $3^{\frac{1}{3}}x$ is on the top and the bottom, so it becomes a one. We are left with a four on the top, and using properties of exponents 4 - 1 = 3, explaining why we have $x^{3}$ leftover too.

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

3 0

2 years ago

Pay ITS TIVE

max2010maxim [7]

Answer:

km per min

Step-by-step explanation:

6km ÷ 60 = 0.1

Bonolo ran 0.1 km per min

it takes 100 mins= 0.1×10

100 mins means 1 hour 40 min

3 0

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,

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