Select the description that is true of the equation y=-5x+5 .

1 answer:

3 0

In y = mx + b form, the slope is in the m position and the y intercept is in the b position.

y = (m)x + (b)
y = -5 x + 5.....slope (m position) = -5....y int (b position) = 5

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Is someone good at geometry i really need help the question doesnt make sense

Nady [450]

M1=[(4a+0)/2 ,(4b+0)/2] = (2a, 2b)
M2=[(4c+4d)/2 ,(4b+0)/2] = [2(c+d) , 2b]

answer
coordinate points (x,y) :
(2a, 2b) , [2(c+d) , 2b]

5 0

3 years ago

An electrician charges $30 for a service call plus $75 per hour of service. If he charged 210$ how many hours did he work

tigry1 [53]

Answer:

2 hours 30 minutes

Step-by-step explanation:

210 - 30 = 190

190 / 75 = 2.5

Sorry if I'm wrong.

4 0

2 years ago

Read 2 more answers

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,

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