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Eddi Din [679]
3 years ago
8

An equation is graphed below. How many solutions does the equation have?

Mathematics
1 answer:
Dimas [21]3 years ago
6 0
One as it intersects only on the Y axis 
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Round 378,903.97 to the nearest thousand
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Answer:

Step-by-step explanation:

378,903,970.

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Find the slope-intercept form of the equation of the line passing through the given points.
Ugo [173]
The\ slope-intercept\ form:y=mx+b\\\\The\ slope-point\ form:y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

(1;\ 1)\to x_1=1;\ y_1=1\\\left(3;-\dfrac{1}{5}\right)\to x_2=3;\ y_2=-\dfrac{1}{5}.\\\\Subtitute:\\\\m=\dfrac{-\frac{1}{5}-1}{3-1}=\dfrac{-1\frac{1}{5}}{2}=-\dfrac{\frac{6}{5}}{2}=-\dfrac{\not6^3}{5}\cdot\dfrac{1}{\not2_1}=-\dfrac{3}{5}


y-1=-\dfrac{3}{5}\left(x-1\right)\\\\y-1=-\dfrac{3}{5}x+\dfrac{3}{5}\ \ \ \ |add\ 1\ to\ both\ sides\\\\\boxed{y=-\frac{3}{5}x+1\frac{3}{5}}


5 0
3 years ago
Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour
UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

C(t) = C(0)e^{rt}

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So C(0) = 110

Husband

Half life of 4 hours. So

C(4) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{4r}

e^{4r} = 0.5

Applying ln to both sides

\ln{e^{4r}} = \ln{0.5}

4r = \ln{0.5}

r = \frac{\ln{0.5}}{4}

r = -0.1733

So for the husband

C(t) = 110e^{-0.1733t}

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.1733t}

C(11) = 110e^{-0.1733*11} = 16.35

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

C(10) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{10}

e^{10r} = 0.5

Applying ln to both sides

\ln{e^{10r}} = \ln{0.5}

10r = \ln{0.5}

r = \frac{\ln{0.5}}{10}

r = -0.0693

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.0693t}

C(11) = 110e^{-0.0693*11} = 51.33

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

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3 years ago
Rewrite the following equation in exponential form.<br> log 49(7) = 1/2
asambeis [7]
It would be log7squared(7)
4 0
3 years ago
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