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3 years ago

(x^2−1)(3x−4)(3x+4) what is the product

Mathematics

1 answer:

Andrei [34K]3 years ago

6 0

Answer:

9x^4-25x^2+16

Step-by-step explanation:

(x^2−1)(3x−4)(3x+4)=(*)

(x^2−1)((3x)^2−4^2) =

(x^2 - 1)(9x^2-16)=

x^2*9x^2 - x^2*16 - 1*9x^2 +16=

9x^4-16x^2-9x^2+16=

9x^4-25x^2+16

(*) (A-B)(A+B) =A^2 - B^2

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A triangle has one side that lies along the line y=1/4x and another that lies along the line y=-1/4x. Which of the following poi

mestny [16]

Answer:

We know that our triangle has one side along the line:

y = (1/4)*x

And other side along the line:

y = -(1/4)*x.

Now, we want to find the vertex.

And we know that the vertex is the point where the two sides conect, so the vertex must be a common point of both lines.

Then we have:

y = (1/4)*x = -(1/4)*x

x = -x

The only solution to that equation is x = 0.

now we evaluate our lines in x = 0 and get:

y = (1/4)*0 = 0

y = -(1/4)*0 = 0

Then the lines intersect in the point (0, 0)

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3 years ago

100(x - 3)=1000 can you help

alekssr [168]

Answer:

x= 13

Step-by-step explanation:

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3 years ago

Read 2 more answers

How many decimals can 2/5 represent?

Alexus [3.1K]

2/5 in decimal form is 0.4.

Hope this helps!

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3 years ago

Five times the sum of a number and 27 is greater than or equal to six times the sum of that

AleksAgata [21]

The complete version of question:

Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.

Answer:

$5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}$

Step-by-step explanation:

As the description of the statement is:

'Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.

Five times the sum of a number and 27 is written as: $5(x + 27)$

greater than or equal is written as: $\geq$

six times the sum of that number and 26' is written as: 6(x + 26)

so lets combine the whole statement:

$5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)$

solving

$5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)$

$5x+135\ge \:6x+156$

$5x+135-135\ge \:6x+156-135$

$5x\ge \:6x+21$

$5x-6x\ge \:6x+21-6x$

$-x\ge \:21$

$\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}$

$\left(-x\right)\left(-1\right)\le \:21\left(-1\right)$

$x\le \:-21$

Therefore,

$5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}$

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3 years ago

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3 years ago