Solve for p if 2|p| = 4

Mathematics

1 answer:

Rina8888 [55]3 years ago

3 0

P is equal to plus or minus 2 because it'll both be equal to 4

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If $a$, $b$, $c$, and $d$ are replaced by four distinct digits from $1$ to $9$, inclusive, then what's the largest possible valu

Alexxx [7]

Answer:

Step-by-step explanation:

for the value of difference to be largest, the minuend should be maximum(most possibly) and the subtrahend should be minimum

[in A-B=X, A is minuend and B is subtrahend ]

so, $a.b should be maximum. as there is a condition that 4 digits should be distinct, the product will be maximum if we choose 2 maximum valued numbers from the given numbers. so, one of them should be 9 and the other should be 8.

therefore, $a.b=9*8=72

as mentioned above, c.d$ should be minimum. this will be possible only when we choose 2 minimum valued numbers from the given numbers. so, one of them should be 1 and the other should be 2.

therefore, c.d$ = 1*2 = 2

hence, the difference = 72-2 = 70

thus, the largest possible value of the difference $a.b - c.d$ = 70

7 0

3 years ago

the hypotenuse of the right angled triangle is 6cm more than twice the shortest side. if the third side is 2 cm less than the hy

Mumz [18]

Answer:

Let the base be p

Hypotenuse = 2p +6

Perpendicular = 2p + 4

By Pythagoras theoram

(2p+6)^2 = (2p+4)^2 +p^2

=> 4p^2 +36 + 24p = 4p^2 + 16 +16p +p^2

=> 36+ 24p = p^2 + 16p + 16

=> p^2 - 8p - 20 = 0

=> p^2 - 10p +2p - 20 = 0

=> p(p-10) +2(p-10) = 0

=> (p-10)(p+2) = 0

p = 10 and - 2

Length can't be negative

So,

p = 10

Base = 10

Perpendicular = 24

Hypotenuse = 26

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3 years ago

How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$