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gregori [183]
2 years ago
8

Solve for b. ab +c=d

Mathematics
2 answers:
pshichka [43]2 years ago
7 0

Answer:

Option 1 is correct.

ab + c = d

b = (d-c) / a

ss7ja [257]2 years ago
6 0

Answer:

The answer is A. b=(d-c)/a

Step-by-step explanation:

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suter [353]
Tan(12) = 40/d
d = 40/tan(12)
d = 188.19m
3 0
3 years ago
Jordan earns $15.35 per hour. on Tuesday , Jordan clocked in at 7:45 Am then clocked out for a break at 12:15 am and working unt
Vaselesa [24]

12:45Answer:

Part A: 9 hours 45 minutes

Part B: $126.1575

Step-by-step explanation:

Part A,

From 7:45 to 12:15 is 4 hours and 30 minutes

From 12:45 which he clocked back in to 6:00 is 5 hours and 15 minutes.

4 hours and 30 minutes plus 5 hours and 15 minutes = 9 hours and 45 minutes.

Part B, 9 hours and 45 minutes can be represented as 9.45

And since he gets 15.35 "per" hour shows we can multiply.

15.35 x 9.45 = 126.1575

Therefor he earned 126.15.

8 0
3 years ago
A bee flies 20 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for fifteen minutes. Then i
Ksenya-84 [330]

A. The equation to find the distance of the flowerbed from the hive is: \frac{x}{20}+\frac{x}{12}+900= 1200 , where x is the distance.

B.  The flowerbed is 2250 feet far from the hive.

<u><em>Explanation</em></u>

Suppose, the distance of the flowerbed from the hive is x feet.

The bee flies 20 feet per second directly to a flowerbed from its hive and flies directly back to the hive at 12 feet per second.

As, Time=\frac{Distance}{Speed}

So, the time taken by the bee to reach the flowerbed = \frac{x}{20} seconds and the time taken to fly back to the hive =\frac{x}{12} seconds.

The bee stays at the flowerbed for 15 minutes or (15×60)seconds or 900 seconds and it is away from the hive for a total of 20 minutes or (20×60)seconds or 1200 seconds.

So, the equation will be.......

\frac{x}{20}+\frac{x}{12}+900= 1200\\ \\ \frac{3x+5x}{60}= 300\\ \\ 8x= 60*300\\ \\ 8x=18000\\ \\ x=\frac{18000}{8}=2250

Thus, the distance of the flowerbed from the hive is 2250 feet.

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.

7 0
3 years ago
Heather has an online jewelry store . She charges $5.85 for shipping on all orders. The expression below can be used to calculat
ArbitrLikvidat [17]

Answer:23.41

Step-by-step explanation:

5 0
3 years ago
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