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3 years ago

"If a = − 9 and b = − 6, show that (a−b) ≠ (b−a)."

Mathematics

1 answer:

denis-greek [22]3 years ago

7 0

Answer:

Step-by-step explanation:

LHS a - b = -9 - (-6) = -9 +6 = -3

RHS b-a = -6 - (- 9) = -6 +9 = 3

as LHS not equal to RHS

a-b not equal to b-a

Thus proven

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Find the values of P for which the quadratic equation 4x²+px+3=0?

klasskru [66]

<h3>Correct Questions :- </h3>

Find the values of P for which the quadratic equation 4x²+px+3=0 , provided that roots are equal or discriminant is zero .

<h3>Solution:- </h3>

Let us Consider a quadratic equation αx² + βx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

For equal roots

D = 0

$\quad\green{ \underline { \boxed{ \sf{Discriminant, D = β² - 4αc}}}}$

So,

$\sf{ β² - 4αc = 0}$

Here,

α = 4
β = p
c = 3

Now,

$\begin{gathered}\implies\quad \sf p²- 4 \times 4 \times 3 =0 \end{gathered}$

$\begin{gathered}\implies\quad \sf p²- 48 =0 \end{gathered}$

$\begin{gathered}\implies\quad \sf p²=48\end{gathered}$

$\begin{gathered}\implies\quad \sf p=±\sqrt{48}\end{gathered}$

$\begin{gathered}\implies\quad \sf p=±\sqrt{2 \times 2 \times 2 \times 2 \times 3} \end{gathered}$

$\begin{gathered}\implies\quad \sf p=± 2\times 2\sqrt{ 3 }\end{gathered}$

$\begin{gathered}\implies\quad \boxed{\sf{p=±4\sqrt{ 3 }}}\end{gathered}$

Thus, the values of P for which the quadratic equation 4x²+px+3=0 are-

4√3 and -4√3.

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2 years ago

For the function f(x) = –5x, which ordered pair will be a point on the graph of the function?

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f(x) = –5x
when x = 0, f(x) = –5(0) = 0

answer is D.
(0, 0)

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3 years ago

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3 years ago

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3 years ago

Read 2 more answers

Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra s

algol [13]

f(x, y) = x ² - 4xy + 5

has critical points where both partial derivatives vanish:

∂f/∂x = 2x - 4y = 0 ==> x = 2y

∂f/∂y = -4x = 0 ==> x = 0 ==> y = 0

The origin does not lie in the region R, so we can ignore this point.

Now check the boundaries:

• x = 1 ==> f (1, y) = 6 - 4y

Then

max{f (1, y) | 0 ≤ y ≤ 2} = 6 when y = 0

max{f (1, y) | 0 ≤ y ≤ 2} = -2 when y = 2

• x = 4 ==> f (4, y) = 12 - 16y

Then

max{f (4, y) | 0 ≤ y ≤ 2} = 12 when y = 0

max{f (4, y) | 0 ≤ y ≤ 2} = -4 when y = 2

• y = 0 ==> f (x, 0) = x ² + 5

Then

max{f (x, 0) | 1 ≤ x ≤ 4} = 21 when x = 4

min{f (x, 0) | 1 ≤ x ≤ 4} = 6 when x = 1

• y = 2 ==> f (x, 2) = x ² - 8x + 5 = (x - 4)² - 11

Then

max{f (x, 2) | 1 ≤ x ≤ 4} = -2 when x = 1

min{f (x, 2) | 1 ≤ x ≤ 4} = -11 when x = 4

So to summarize, we found

max{f(x, y) | 1 ≤ x ≤ 4, 0 ≤ y ≤ 2} = 21 at (x, y) = (4, 0)

min{f(x, y) | 1 ≤ x ≤ 4, 0 ≤ y ≤ 2} = -11 at (x, y) = (4, 2)

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3 years ago