All categories

3 years ago

Factor 15abc − 9ab − 3bcd completely

Mathematics

2 answers:

love history [14]3 years ago

8 0

the correct Answer is d

Step-by-step explanation:

goldenfox [79]3 years ago

3 0

I believe the correct answer is C

You might be interested in

I need help with this problem!!! It’s confusing me so much and I need it by tomorrow!please tell me the answer and how you got i

ExtremeBDS [4]

I think its 3 - (0,10, 42)

if you put in the values of the given x, then these are the resulting y values.

8 0

3 years ago

Find the area of a floor which is covered by 1050 tiles.each tiles measuring 20 square meters

RoseWind [281]

Answer:

area of floor is

1050*20=21000

Step-by-step explanation:

mark me as brainliest

7 0

2 years ago

Will give brainliest if right, please help me with my homework

Rus_ich [418]

Answer:

Step-by-step explanation:

6 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

Help anyone please i need it

zheka24 [161]

Answer:

The sum is equal to $-905,200$

Step-by-step explanation:

we know that

The formula to calculate the sum is equal to

$Sn=n\frac{(a1+an)}{2}$

In this problem

$n=775$ -----> number of terms

a1 is the first term

$a1=(-3(1)-4)=-7$

an is the last term

$an=(-3(775)-4)=-2,329$

substitute

$Sn=775\frac{(-7-2,329)}{2}=-905,200$

4 0

3 years ago