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andriy [413]
2 years ago
6

Please help!!! Which of the following expand (x + y)^5 using Pascal’s Triangle?

Mathematics
1 answer:
pishuonlain [190]2 years ago
6 0

Answer:

{x}^{5}  + 5 {x}^{4} y + 10 {x}^{3}  {y}^{2}  + 10 {x}^{2}  {y}^{3}  + 5x {y}^{4}  +  {y}^{5}

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A student said that since –9 is less than 4, then |–9| is less than |4|. Is the student correct? Explain why or why not
Leno4ka [110]
The student is not correct.

Absolute value means distance from zero
|-9| = 9
|4| = 4

Since 9 is greater than 4,
|-9| is greater than |4|

In other words, -9 is further from zero then 4
5 0
3 years ago
What is the domain and range of the relation shown in the table?
Leokris [45]

Answer:

domain: {-12, -8, 0, 1}   range: {0, 8, 12}

Step-by-step explanation:

domain are of all the input values shown on the x-axis. The range is the set of possible output  shown on the y-axis.

5 0
3 years ago
(PLEASE HELP) The histogram shows the number of hours that 25 college students spent studying for their final exams. Which state
Semmy [17]

The statement on the bottom left is correct. (6 students studied for 20 to 24 hours)

6 0
3 years ago
Read 2 more answers
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
3 years ago
2.5g + 3.14g + 6g + 12.32g
klio [65]

Answer:

23.96g

Step-by-step explanation:

Make sure to thank me!!!

7 0
3 years ago
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