All categories

3 years ago

Subtract 3a^2+5ab−4b^2 from 8a^2−6ab+4b^2

Mathematics

1 answer:

liberstina [14]3 years ago

3 0

Answer:

$\large\boxed{(8a^2-6ab+4b^2)-(3a^2+5ab-4b^2)=5a^2-11ab+8b^2}$

Step-by-step explanation:

$(8a^2-6ab+4b^2)-(3a^2+5ab-4b^2)\\\\=8a^2-6ab+4b^2-3a^2-5ab-(-4b^2)\\\\=8a^2-6ab+4b^2-3a^2-5ab+4b^2\qquad\text{combine like terms}\\\\=(8a^2-3a^2)+(-6ab-5ab)+(4b^2+4b^2)\\\\=5a^2-11ab+8b^2$

You might be interested in

Find the ratio in the simplest form 15 minutes to 2 hours

Lelu [443]

15min:2h
15min:120min
15/120
3/24
1:8 is the answer

3 0

3 years ago

B is the midpoint of ac. bc = 3x - 1. ac = x + 3. find the value of ab

Ilia_Sergeevich [38]

Answer:

ab=bc because b devide ac in two equal sizes

ac=ab+bc

ac=2×ab

ac =2×bc

lets substitute ac and bc there we are given

x+3=2(3x-1)

x+3=6x-2

-5x/-5=-5/-5

x=1

ab=3(1)-1

ab=2

8 0

2 years ago

Michelle is planning an anniversary party at a restaurant . There is a fee to use the banquet room , plus the cost of dinner eac

KonstantinChe [14]

Answer:c

Step-by-step explanation: The party will be grand

3 0

3 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

Enter the y coordinate of the solution to this system of equations. <br> -2x+3y=-6<br> 5x-6y=15

julia-pushkina [17]

Y coordinate on solving both equations comes out to be 0
-2x+3y=-6
3y = -6+2x
put the value of 3y in equation 2nd
5x-2(-6+2x) =15
5x+12-4x = 15
x=3
put value of X in 3y = -6+2x
3y = -6+2*3 = -6+6 = 0
thus y = 0

7 0

3 years ago