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2 years ago

The polynomial x^3+ px^2-x-2 has x-1 as a factor. what is the value of p?

Mathematics

1 answer:

Archy [21]2 years ago

5 0

Comparing x-1 with x-a then x=1
1^3+p*1^2-1-2=0
1+2p-3=0
2p=3-1
P=2/2
P=1
:.p=1

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A store had 150 laptops in the month of January. Every month, 20% of the laptops were sold and 10 new laptops were stocked in th

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<u>Answer-</u>

$\boxed{\boxed{f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0}}$

<u>Solution-</u>

Here, n represents the number of months and f(n) represents the number of laptops in the store after n months.

As the store had 150 laptops in the month of January or at the beginning.

So $f(0)=150$

Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.

As 20% of laptops were sold, so 80% were in the store.

So, after one month total number of laptops in the store,

$\Rightarrow f(1)=0.8(150)+10$

$\Rightarrow f(1)=0.8(f(0))+10$ $(\because f(0)=150)$

Again after one month total number of laptops in the store,

$\Rightarrow f(2)=0.8\times (0.8(150)+10)+10$

$\Rightarrow f(2)=0.8(f(1))+10$ $(\because f(1)=0.8(150)+10)$

Analyzing the pattern, the recursive function f(n) will be,

$f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0$

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2 years ago

Read 2 more answers

Please help me on this

Readme [11.4K]

For this case we have to define root properties:

$\sqrt [n] {a ^ n} = a ^ {\frac {n} {n}} = a$

In addition, we know that:

$a ^ {- 1} = \frac {1} {a ^ 1} = \frac {1} {a}$

On the other hand:

$4 ^ 2 = 16$

Thus, we can rewrite the given expression as:

$\sqrt {4 ^ 2 * a ^ 8 * \frac {1} {b ^ 2}} =\\4 ^ {\frac {2} {2}} * a ^ {\frac {8} {2}} * \frac {1} {b ^ {\frac {2} {2}}} =\\4 * a ^ 4 * \frac {1} {b}$

ANswer:

Option B

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