All categories

3 years ago

Credit cards can only lead to debt that is difficult to eliminate and should be avoided at all costs. True or false?

Mathematics

2 answers:

Ulleksa [173]3 years ago

7 0

Answer:

The given statement is false.

Step-by-step explanation:

Credit cards can only lead to debt that is difficult to eliminate and should be avoided at all costs.

This statement is false.

This is not a generalization that credit cards always lead to debts. Rather misuse of credit cards and not paying the bills timely, can lead to debt.

Valentin [98]3 years ago

5 0

I think It’s false because not everyone has a hard time to eliminate debts

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Vinil7 [7]

Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

$((x+y)-5)(2(x+y+1)$

$((x+y)-5)(2x+2y+1)$

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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One of the great mysteries of the world is my hotdogs and hotdog buns come in packages of different sizes hotdogs come in packs

V125BC [204]

Answer:

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Step-by-step explanation:

hot dogs come in packs of 8

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