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2 years ago

Written as a simplified polynomial in standard form, what is the result when (2x - 7) ^ 2 is subtracted from 1?

Mathematics

1 answer:

zaharov [31]2 years ago

8 0

Answer:

ans: -4( x-3 )( x-4 ) I.e -4( x² + x - 12 )

Step-by-step explanation:

as per question

1- (2x-7)²

= (1 + 2x - 7)(1- 2x + 7)

= ( 2x - 6)(-2x +8)

= -4(x-3)(x+4)

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Factor x^2 + 5x + 10 completley

k0ka [10]

This should be the answer.

5 0

3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

2 years ago

Solve by substitution. Show each step of your work.<br><br> 2x + y = 7<br><br> 3x + 5y = 14

Monica [59]

Alright...simple...showing all steps.. ;)

You have the equations...

2x+y=7
and
3x+5y=14

To be able to even solve for any of the variables...multiply the equations by...2...and..3...

2x+y=7----*3--> 6x+3y=21
and
3x+5y=14-----*2--->6x+10y=28

Thus,

6x+3y=21
-
6x+10y=28

=========

-7y=-7

y=1

Now, plug y back into any of the original equations....we'll use 2x+y=7 in this case....

2x+(1)=7

2x+1=7
-1 -1

2x=6

x=3

Thus, the point of intersection for these two equations is (3,1)

8 0

3 years ago

Can a rational number be subtracted by a irrational number an equal a rational number?

vladimir2022 [97]

Answer:

Step-by-step explanation:

The sum of a rational number and an irrational number is irrational.

4 0

2 years ago

Right or wrong ????

tangare [24]

Answer:

c 1/2

Step-by-step explanation:

You are incorrect.

We are going from ABCDEFGH to A'B'C'D'E'F'G'H'. We are going from the large octagon to the small octagon. The scale factor of a dilation is the number you multiply the original lengths to get the lengths of the dilation.

$scale~factor = \dfrac{length~of~dimension~in~dilated~image}{length~of~corresponding~dimension~in~original~figure}$

Look at point (6, 0) which becomes (3, 0) in the dilation.

The distance from the origin to (6, 0) is 6 units.

The distance from the origin to (3, 0) is 3 units.

scale factor = dilated/original = 3/6 = 1/2

The scale factor of the dilation is 1/2.

Also look at segment AB whose length is 4. The length of segment A'B' is 2.

scale factor = dilated/original = 2/4 = 1/2

Answer: 1/2

4 0

3 years ago