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2 years ago

Which figure correctly demonstrates using a straight line to determine that the graphed equation is not a function of x?

Mathematics

2 answers:

Ulleksa [173]2 years ago

3 0

Answer:

Lower left.

Step-by-step explanation:

To determine if a graph is a function the "vertical line test is used. If any point on a graph has one x value go to two y values it is not a function.

Witht he four pictures, the two on the right use a horizontal line, so it doesn't tell at all, on the leftthe upper one is vertical but it doesn't show that one x value has two corresponding y values. So it must be the lower left one.

The lower left picture has a line at x=1, and it passes through the line of the graph at two different y values, 4 and -4. So this has two y values with one x value.

Semmy [17]2 years ago

3 0

Answer:

option C

Step-by-step explanation:

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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.

7 0

3 years ago

Greg swam 4 kilometers against the current in the same amount of time it took him to swim 16 kilometers with the current. The ra

ad-work [718]

Answer:

5 kmph

Step-by-step explanation:

Given: Greg swam 4 kilometers against the current.

Greg swam 16 kilometers with the current.

Rate of the current was 3kmph.

Lets assume the speed of greg swimming with no current be "x".

∴ Speed of swimming against current= $(x-3)\ kmph$

Speed of swimming with current= $(x+3)\ kmph$

As given, it took same amount of time to swim both with current and against current.

∴ Forming an equation to find the value of x, which is speed of greg swimming with no current.

We know, $Time= \frac{Distance}{Speed}$

⇒ $\frac{4}{(x-3)} = \frac{16}{(x+3)}$

Multiplying both side by (x+3) and (x-3)

⇒ $4\times (x+3)= 16\times (x-3)$

Using distributive property of multiplication.

⇒ $4x+12= 16x-48$

Subtracting both side by 4x and adding by 48.

⇒ $60= 12x$

Dividing both side by 12

⇒ $x= \frac{60}{12} = 5\ kmph$

Hence, Greg can swim at the rate of 5kmph with no current.

7 0

3 years ago

Plz help me I will mark u brainiest

Likurg_2 [28]

Answer:

quadrant one

Hope this helps

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3 years ago

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loris [4]

I divided 56 by 4 to see how much per CD Brad payed. This got me $14. $42 divided by $14 is $3, so he can buy 3 CDs.

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2 years ago

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Flura [38]

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3 years ago

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