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Kay [80]
3 years ago
8

Which points are possible approximations for this system? Select two options. (1.9, 2.5) (2.2, –1.4) (2.2, –1.35) (1.9, 2,2) (1.

9, 1.5)
Mathematics
2 answers:
Rudiy273 years ago
4 0

Answer:

second: (2.2,-1.4)

third: (2.2,-1.35)

Step-by-step explanation:

svet-max [94.6K]3 years ago
4 0

Answer:

2nd and 3rd

Step-by-step explanation:

                                             

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A company pays a bonus to four employees A, B, C, and D. A gets four times as much as B. B gets 50% of the amount paid to C. C a
Tomtit [17]

Answer:

A = 800, B = 200, C = 400 Andy D = 400

Step-by-step explanation:

5 0
3 years ago
The middle school ski club plans ski trips and participants in ski competitions. If 25% of the students participate in ski compe
KonstantinChe [14]

Answer:

7/9

Step-by-step explanation:

6 0
3 years ago
Number 34 please guys!!!!!!
GenaCL600 [577]
The answer is obviously three.

3 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.

7 0
3 years ago
Tbh i’m too lazy to do it‍♀️
rusak2 [61]

Answer:

115.56 but since you have to round to the nearest km is 116

Step-by-step explanation:

3 0
3 years ago
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