All categories

3 years ago

Expand the binomial (2x + y2)5. The fourth term in the expansion of the binomial (2x + y2)5 is

Mathematics

2 answers:

ankoles [38]3 years ago

5 0

Answer:

Expanded form: $32x^{5} + 160x^{4}y + 320x^{3}y^{2} + 320x^{2}y^{3} + 160xy^{4} + 32y^{5}$

Fourth term: 320x^{2}y^{3}

Step-by-step explanation:

The expansion is given by the following rule:

$(nx + my)^{5} = (nx)^{5} + 5(nx^{4})(my) + 10(nx)^{3}(my)^{2} + 10(nx)^{2}(my)^{3} + 5(nx)(my^{4}) + (my)^{5}$

Given this we just need to replace the values such as n = 2 and m = 2 for the binomial shown.

For the fourth term we just need to count after we have the expanded form and then we have our answer.

Ainat [17]3 years ago

3 0

Thank you for posting your question here at brainly. The fourth term in the expansion of the binomial (2x + y2)5 is equal to 2x5 + 160x4y + 320x3y2 + 320x2y3 + 160xy4 + 32y5. I hope the answer will help you.

You might be interested in

in a random sample of 150 voters, 60 support Trump. Set up a 96 percent confidence interval for the proportion of voters who sup

denis-greek [22]

Given:
n = 150, sample size
Denote the sample proportion by q (normally written as $\hat{p}$ ).
That is,
q = 60/150 = 0.4, sample proportion.

At the 96% confidence level, the z* multiplier is about 2.082, and the confidence interval for the population proportion is
$q \pm z^{*}[ \frac{q(1-q)}{ \sqrt{n} } ]$

That is,
0.4 +/- 2.082* √[(0.4*0.6)/150]
= 0.4 +/- 0.0833
= (0.3167, 0.4833)
= (31.7%, 48.3%)

Answer: The 96% confidence interval is about (32% to 48%)

3 0

3 years ago

Three consecutive integers have a sum of 234. What are the three integers?

Wewaii [24]

X+x+1+x+2 = 234
3x + 3 = 234
3x = 231
x = 77

so Three consecutive integers: 77,78,79

double check
77+78+79=234

5 0

3 years ago

Solve the equation for y. Then find the value of y for each value of x. 3x-7y=13;x= -3, 0, 3

Sidana [21]

When x is -3

$3\left(-3\right)-7y=13\quad :\quad y=-\frac{22}{7}\quad \left(\mathrm{Decimal}:\quad y=-3.14285\dots \right)$

When x is 0

$3\cdot \:0-7y=13\quad :\quad y=-\frac{13}{7}\quad \left(\mathrm{Decimal}:\quad y=-1.85714\dots \right)$

When x is 3

$3\cdot \:3-7y=13\quad :\quad y=-\frac{4}{7}\quad \left(\mathrm{Decimal}:\quad y=-0.57142\dots \right)$

7 0

3 years ago

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

System Of Linear Equations

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago

Help ASAP please again and again

AnnyKZ [126]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers