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3 years ago

Are the following lines parallel, perpendicular, or neither?

Mathematics

1 answer:

sashaice [31]3 years ago

7 0

Answer:

Neither

Step-by-step explanation:

equation 1 is a negative slope, equation 2 is a positive slope, but with a different slope.

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What should be added to a²-2ab+b² to get a²-ab-2b²

dybincka [34]

SOLUTION:

what should be added to a²-2ab+b² to get a²-ab-2b²

let the added no is P

$\pmb{P+(a^2-2ab+b^2)=(a^2-ab-2b^2) }$

$\pmb{P=(a^2-ab-2b^2)-(a^2-2ab+b^2) }$

$\pmb{P=a^2-ab-2b^2-a^2+2ab-b^2 }$

$\pmb{P=ab-3b^2 }$

$\therefore{\pmb{ab-3b^2 }}$ should be added to a²-2ab+b² to get a²-ab-2b²

5 0

3 years ago

Find the value of x number on top is 6 and other is 9

faltersainse [42]

Answer:

x = - 12, 3

Step-by-step explanation:

By tangent secant theorem:

$x(x + 9) = {6}^{2} \\ \\ {x}^{2} + 9x = 36 \\ \\ {x}^{2} + 9x - 36 = 0 \\ \\ {x}^{2} + 12x - 3x - 36 = 0 \\ \\ x(x + 12) - 3(x + 12) = 0 \\ \\ (x + 12)(x - 3) = 0 \\ \\ x + 12 = 0, \: \: x - 3 = 0 \\ \\ x = - 12, \: \: x = 3 \\ \\$

5 0

2 years ago

How do i do this help

Ratling [72]

I don't know if u can see it, but there.
X=12
Y=15
Z=20

5 0

3 years ago

Find the value of each of the following

Mandarinka [93]

Answer: C

Step-by-step explanation:

8 0

3 years ago

What is the cost of operating a 3.00-w electric clock for a year if the cost of electricity is $0.0900 per kw · h ?

brilliants [131]

We must take into account the following change of units:
$1kw = 1000w

1 year = 365 days

1 day = 24 hours$
Applying the change of units we have that the electric consumption for 1 year is given by:
$(3) * (1/1000) * (365) * (24) = 26.28 kw * h
$
Then, the total cost is given by:
$(26.28) * (0.0900) = 2.3652
$
Answer:
the cost of operating a 3.00-w electric clock for a year is:
$ 2.3652

8 0

3 years ago