All categories

4 years ago

Is this right? Can someone tell me

Mathematics

1 answer:

BARSIC [14]4 years ago

6 0

Answer:

Step-by-step explanation:

m<1 = m<2

8y - 6 = 7y

8y - 7y = 6

y = 6°

m<2 = 7y = 7(6) = 42°

You might be interested in

Can 120/27 be simplified

svlad2 [7]

yes, the most simplified version would be 40/9

in decimal form this is 4.444444

8 0

4 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

Which graph represents the equation y-2=3x

Naily [24]

Answer:

to be able to answer that for you we would need graphs but the graph should have these three points -2,-7 0,2 and 2,8 it will also be a diagonal line going up right

Step-by-step explanation:

5 0

4 years ago

How do I solve 3(2x+1)

jonny [76]

to solve 3(2x+1) you need to multiply. so 3 times 2x is 6x and 3 times 1 is 3 making the final answer to be 6x+1

6 0

4 years ago

Read 2 more answers

6.<br> Chapter 6 Quiz 2<br> (Lessons 6-3 and 6-4)

Firlakuza [10]

Answer:

What in the world is this even for?

5 0

3 years ago