All categories

4 years ago

Which of the following products is greater than 6. A: 6×2/3. B: 6× 1/3. C: 6× 1 1/6. D: 6× 1/2 Don't mind all random work

Mathematics

1 answer:

Colt1911 [192]4 years ago

3 0

Answer:

C. 6×11/6

Step-by-step explanation:

Because 11/6 is greater than 1, if you were to multiply it by 6, you would get a product greater than 6.

You might be interested in

Solve for x in both!!!!!

zvonat [6]

Answer:

$- 2(x - 2) - 4x = 3(x + 1) - 9x \\ - 2x + 4 - 4x = 3x + 3 - 9x \\ 9x - 3x - 2x - 4x = 3 - 4 \\ 9x - 9x = - 1 \\ 0x = - 1 \\ 0 = - 1$

So ; No solution

___o___o____

$5(x + 2) - 3 = 3x - 8x + 7 \\ 5x + 10 - 3 = - 5x + 7 \\ 5x + 7 = - 5x + 7 \\ 5x + 5x = 7 - 7 \\ 10x = 0 \\ \\ x = \frac{0}{10} \\ \\ x = 0$

I hope I helped you^_^

4 0

3 years ago

Read 2 more answers

A theater sells adult tickets for $8.00 and children's tickets for $5.50. Sally bought some of each kind (not the same number) a

Setler79 [48]

D is correct answer. X stands for adult tickets, y for children

8 0

4 years ago

Read 2 more answers

Kary needs 54 licks for every 3 lollipops she eats. Which option below will complete the table?

finlep [7]

Answer:

The second one is correct as 108 is double 54

Step-by-step explanation:

Therefore since 54 licks gets one lollipop gone 108 would get 2 lollipops gone

4 0

3 years ago

At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37

monitta

Answer:

Half-life of the goo is 49.5 minutes

$G(t)= 300e^{-0.014t}$

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let $M_0\,,\,M_f$ denotes initial and final mass.

$M_0=300\,\,grams\,,\,M_f=37.5\,\,grams$

According to exponential decay,

$\ln \left ( \frac{M_f}{M_0} \right )=-kt$

Here, t denotes time and k denotes decay constant.

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014$

So, half-life of the goo in minutes is calculated as follows:

$\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes$

Half-life of the goo is 49.5 minutes

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}$

So,

$G(t)= M_f=M_0e^{-kt}$

Put $M_0=300\,\,grams\,,\,k=0.014$

$G(t)= 300e^{-0.014t}$

Put t = 32 minutes

$G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams$

7 0

3 years ago

What is the slope of this line?

DaniilM [7]

Answer:

Step-by-step explanation:

5 0

3 years ago