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3 years ago

Find the product of 2x4(4x2 + 3x + 1).

Mathematics

2 answers:

tresset_1 [31]3 years ago

4 0

$2x^4(4x^2 + 3x + 1)=\\\\(2x^4)(4x^2)+(2x^4)(3x)+(2x^4)(1)=\\\\8x^6 + 6x^5+2x^4$

The answer is A.

Bond [772]3 years ago

4 0

Answer:

The product of the $2x^{4}(4x^{2}+3x+1)$ is $8x^{6}+6x^{5}+2x^{4}$ .

Step-by-step explanation:

As given the expression in the question be

$= 2x^{4}(4x^{2}+3x+1)$

First open the bracket

$= 2x^{4}\times 4x^{2}+2x^{4}\times 3x+2x^{4}\times 1$

Now by using the property

$x^{a}\times x^{b}=x^{a+b}$

$= 8x^{6}+6x^{5}+2x^{4}$

Therefore the product of the $2x^{4}(4x^{2}+3x+1)$ is $8x^{6}+6x^{5}+2x^{4}$ .

You might be interested in

Find the distance between the point - 3, 2 and 1 - 1

dusya [7]

Answer:

5 units

Step-by-step explanation:

Find the distance between the point - 3, 2 and 1 - 1

Given data

x1= -3

y1= 2

x2= 1

y2= -1

The expression for the distance between two points is

d=√((x_2-x_1)²+(y_2-y_1)²)

subtitute

d=√((1-(-3))²+(-1-(2))²)

d=√((1+3))²+(-1-2))²)

d=√((4)²+(-3))²)

d=√16+9

d=√25

d=5

Hence the distance between the points is 5 units

7 0

2 years ago

A group of students were given a spelling test.

Yuki888 [10]

Answer:

a) 2

b)33

c)15

hope this helps you!

5 0

2 years ago

How can I solve this ?

LuckyWell [14K]

Answer:

985.25

Step-by-step explanation:

An easy way to divide by larger numbers is to see if the divisor fits into the thousands, hundreds, tens, and ones place.

8 doesn't go into 7, so you'd bring down the number in the hundreds place and so on and so forth.

3 0

3 years ago

Read 2 more answers

If 10-10ˣ=9.9, find x

madam [21]

we are given

$10-10^x=9.9$

we can move 10 on right side

$-10^x=9.9-10$

$-10^x=-0.1$

we cancel -1

$10^x=0.1$

we can rewrite right side

$10^x=10^{-1}$

we can compare exponents

and we get

$x=-1$ .............Answer

3 0

3 years ago

At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon o

maria [59]

Answer:

a) $s(t) =400- 385 e^{-\frac{1}{10} t}$

b) $s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

$Rate change = In-Out$

And we can name the rate of change as $\frac{ds}{dt}=rate change$

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is $\frac{s}{50gal}$

And now we can find the expression for the amount of salt after time t like this:

$\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}$

And we have this differential equation:

$\frac{dS}{dt} +\frac{1}{10} s = 40$

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

$\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}$

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

$e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt$

$e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C$

Now we can divide both sides by $e^{\frac{1}{25} t}$ and we got:

$s(t) =400 + C e^{-\frac{1}{10} t}$

Now we can apply the initial condition in order to solve for the constant C like this:

$15 = 400+C$

$C=-385$

And then our function would be given by:

$s(t) =400- 385 e^{-\frac{1}{10} t}$

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

$s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

7 0

2 years ago