All categories

2 years ago

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:

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

You might be interested in

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