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solong [7]
4 years ago
14

How the hell is the answer 6 or -4

Mathematics
1 answer:
hram777 [196]4 years ago
3 0

Answer:

The values of y is -4 and 6

Step-by-step explanation:

Solve by factoring y^2 -2y=24

We need to factorize the term y^2 -2y=24.

Rearranging:

y^2 - 2y -24 =0

For factorizing we need to break the middle term such that their sum is equal to -2y and product is equal to -24y^2

We know, -6y*+4y = 24y^2 and -6y + 4y = -2y

y^2 -6y +4y -24 =0

y(y-6) +4(y-6) = 0

(y+4)(y-6)=0

Now, y+4 =0 and y-6 =0

Finding value of y,

y = -4 and y =6

So, the values of y is -4 and 6

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8 0
4 years ago
Which equation represents the statement? mark's age, x, is 6 times his age 2 years ago.
s2008m [1.1K]
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3 0
3 years ago
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Step-by-step explanation:

It looks like it would be c=4b but 4*7 equals 28 not 30

4 0
3 years ago
There are currently 441 dairy cows at Dancing Dairy Farm. Due to some considerations, the number of dairy cows is decreasing at
Ymorist [56]

Answer:

the production of milk at the Dancing Dairy Farm is increasing by 2158 gallons/year

Step-by-step explanation:

Given the data in the question;

Let us represent the number of cows at the farm with x and

each cow produces y gallons of milk

Total mil production will be T.

so

T = x × y

now, differentiating with respect to t

dT/dt = d( xy )/dt

dT/dt = xdy/dt + ydx/dt

given that;

x = 441

y = 1157

dx/dt = -13

dy/dt = 39

so we substitute

dT/dt = ( 441 )( 39 ) + ( 1157 )( -13 )

dT/dt = 17199 - 15041

dt/dT = 2158

Therefore, the production of milk at the Dancing Dairy Farm is increasing by 2158 gallons/year

3 0
3 years ago
Show that 3x^2 -2x + 1 is always greater than 0.<br> ( This is an Additional Math Question )
Leviafan [203]

Answer:

3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3

(x-1/3)^2 is greater or equal to zero

3(x-1/3)^2 is greater or equal to zero

and 2/3 is greater than zero

So there sum is greater than zero

Proved

Step-by-step explanation:

3x^2 -2x + 1 =3(x^2-2/3x+1/3)

Consider x^2-2/3x+1/3

Remember that (a-b)^2 =a^2-2ab+b^2

x^2=a^2

a=x

-2/3x= -2*x*b

b=1/3

S0 (x-1/3)^2= x^2-2/3x+1/9

x^2-2/3x+1/3= x^2-2/3x+1/9+1/3-1/9= (x-1/3)^2+2/9

3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3

(x-1/3)^2 is greater or equal to zero

3(x-1/3)^2 is greater or equal to zero

and 2/3 is greater than zero

So there sum is greater than zero

Proved

6 0
3 years ago
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