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3 years ago

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Julius sells two robots a small one and a larger model. the price for the smaller robot is X and the price for the large robot i

s Y. write an equation that will let him earn a profit of $13,050 if he sells 15 small robots and 42 large robots.

1 answer:

Mademuasel [1]3 years ago

8 0

Answer:

15x + 42y = $13 050

Step-by-step explanation:

hope that helped :)

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Yolanda is planning to put a pool in her backyard. The scaled model below gives the reduced measures for diameter and depth. The

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Answer:

C option: 6 feet

Step-by-step explanation:

Since Yolanda wants to keep the pool in proportion to the model, the ratio of diameter to depth of model and the pool will be same.

Let the depth of pool is x feet.

Ratio of Diameter to Depth of Model = Ratio of Diameter to Depth of pool

This means the depth of pool should be 6 feet if Yolanda wants to keep the pool in proportion to the model.

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Step-by-step explanation:

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What is the value of x<br><br> Triangle with sides 10 6 x-6

Grace [21]

Given that the triangle is a right angled triangle, to solve for x we shall use the Pythagorean theorem given by:
c^2=a^2+b^2
where:
c is the hypotenuse
a and b are the legs.
c=10
a=6
b=(x-6)
thus plugging the values in the equation and solving for x we get:
10^2=6^2+(x-6)^2
100=36+x^2-12x+36
100=x^2-12x+72
0=x^2-12x-28
factoring the above we get:
0=(x+2)(x-14)
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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,

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

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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