All categories

4 years ago

Raj was asked to fully simplify this polynomial and put it into standard form.

Mathematics

2 answers:

tatuchka [14]4 years ago

8 0

Answer:

6x³

Step-by-step explanation:

2x²y + 8x³ – xy² – 2x³ + 3xy² + 6y³=

6x³+2x²y+2xy²+6y³

natita [175]4 years ago

3 0

Step-by-step explanation:

The first term is 6x3

Because he subtracted like terms 8x3 and 2x3

You might be interested in

Trevor pays $125 in fees each semester plus $650 per college course.

slava [35]

I hope this helps you

for per college course $650

650.?+125=2725

7 0

3 years ago

HELP!! ILL MARK BRAINLIEST

Sonja [21]

Answer:

Step-by-step explanation:

youre welcome

7 0

3 years ago

Read 2 more answers

The vertices of a square are A(–2, 4), B(4, 4), C(4, –2), and D(–2, –2). The diagonals of the square intersect at their midpoint

sertanlavr [38]

A(–2, 4), B(4, 4), C(4, –2), D(–2, –2)
-> select some diagonal: AC
midpoint is at A+(1/2)*AC

-> calculate vector AC:
AC=(4-(-2),-2-4)=(6,-6)
-> calculate half length of AC:
(1/2)*AC=(3,-3)

A+(1/2)*AC=
(-2,4)+(3,-3)=(1,1) which is the midpoint

so the first option is correct

4 0

3 years ago

Which algebraic expression represents the phrase "four times a number"?

Delvig [45]

Answer:

O 40

Step-by-step explanation:

because if you are doing algebra the term for a number is a letter and instead of doing 4xO you substituted to 4O

7 0

3 years ago

The line segment AB with endpoints A (-3, 6) and B (9, 12) is dilated with a scale

Gwar [14]

Answer:

C) (-2, 4), (6,8) is the correct answer.

Step-by-step explanation:

Given that line segment AB:

A (-3, 6) and B (9, 12) is dilated with a scale factor 2/3 about the origin.

First of all, let us calculate the distance AB using the distance formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here,

$x_2=9\\x_1=-3\\y_2=12\\y_1=6$

Putting all the values and finding AB:

$AB = \sqrt{(9-(-3))^2+(12-6)^2}\\\Rightarrow AB = \sqrt{(12)^2+(6)^2}\\\Rightarrow AB = \sqrt{144+36}\\\Rightarrow AB = \sqrt{180}\\\Rightarrow AB = 6\sqrt{5}\ units$

It is given that AB is dilated with a scale factor of $\frac{2}{3}$ .

$x_2'=\dfrac{2}{3}\times x_2=\dfrac{2}{3}\times9=6\\x_1'=\dfrac{2}{3}\times x_1=\dfrac{2}{3}\times-3=-2\\y_2'=\dfrac{2}{3}\times y_2=\dfrac{2}{3}\times 12=8\\y_1'=\dfrac{2}{3}\times y_1=\dfrac{2}{3}\times 6=4$

So, the new coordinates are A'(-2,4) and B'(6,8).

Verifying this by calculating the distance A'B':

$A'B' = \sqrt{(6-(-2))^2+(8-4)^2}\\\Rightarrow A'B' = \sqrt{(8)^2+(4)^2}\\\Rightarrow A'B' = \sqrt{64+16}\\\Rightarrow A'B' = \sqrt{80}\\\Rightarrow A'B' = 4\sqrt{5}\ units = \dfrac{2}{3}\times AB$

So, option C) (-2, 4), (6,8) is the correct answer.

5 0

3 years ago