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stira [4]
3 years ago
7

Michelle wants to carpet a room that is 215 square feet. The chart below indicates

Mathematics
1 answer:
zimovet [89]3 years ago
4 0
Explained in the photo below. Hope this is right!

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PLS HELP ME GIVING BRAINLIEST!!!!
Wewaii [24]

Answer:

Brainly patrol stop deleting my answers

A. 2

B. 4

C. 3

D. 7

Step-by-step exp

7 0
3 years ago
Read 2 more answers
how many 1 1/2 centimeter cubes can fit into a rectangular prism that has a length of 12 centimeters , a width of 6 centimeters
VLD [36.1K]

The long, straightforward, brute-force way to slog through the problem:

The prism is 12cm x 6cm x 9cm.

Volume of the prism = (12 x 6 x 9) = 648 cm³

Volume of each little cube = (1.5 x 1.5 x 1.5) = 3/375 cm³

Number of little cubes that fit into the prism =  (648 cm³) / (3.375 cm³) = <em>

                                                                                     192 of them</em>


=============================================

The elegant but almost equally long way to master the problem:

The prism is 12cm x 6cm x 9cm.

If each dimension of your measuring cubie is 1.5cm,
then the prism measures

     (8 cubie lengths) x (4 cubie widths) x (6 cubie heights) .

Its volume is  (8 x 4 x 6) = <em>192 measuring cubies</em>.


8 0
3 years ago
Can you evaluate 6x-3y-3x?
Hunter-Best [27]

Answer:

= 3x - 3y

Step-by-step explanation:

7 0
3 years ago
Help me please :( i really need help
iris [78.8K]

Answer:

B)41.2

C)0.736

D)6.132

Step-by-step explanation:

B) 412/10=41.2

C)73.6/100=0.736

D)6132/1000=6.132

hope this helps

6 0
3 years ago
Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou
mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

\frac{dQ}{dt} = -kQ

In which k is the decay rate.

The solution is:

Q(t) = Q(0)e^{-kt}

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that Q(0) = 8000

500 counts per minute 13 days later.

This means that Q(13) = 500. We use this to find k.

Q(t) = Q(0)e^{-kt}

500 = 8000e^{-13k}

e^{-13k} = \frac{500}{8000}

\ln{e^{-13k}} = \ln{\frac{500}{8000}}

-13k = \ln{\frac{500}{8000}}

k = -\frac{\ln{\frac{500}{8000}}}{13}

k = 0.2133

So

Q(t) = Q(0)e^{-0.2133t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.2133t}

0.5Q(0) = Q(0)e^{-0.2133t}

e^{-0.2133t} = 0.5

\ln{e^{-0.2133t}} = \ln{0.5}

-0.2133t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.2133}

t = 3.25

The half-life of the radioactive substance is of 3.25 days.

7 0
3 years ago
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