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3 years ago

Divide..................

Mathematics

1 answer:

Masja [62]3 years ago

6 0

Answer:

option-A

Step-by-step explanation:

we are given

Dividend is

$6x^2-19x+10$

Divisor is

$2x-5$

we can use long division method

we get

Remainder is

Quotient is

$3x-2$

so, we can write as

$\frac{6x^2-19x+10}{2x-5}=3x-2+\frac{0}{2x-5}$

$\frac{6x^2-19x+10}{2x-5}=3x-2$

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3 years ago

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Answer:

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3 years ago

A box contains 8 two-inch screws. Four have a Phillips head and 4 have a slotted head. In how many ways can 4 screws be chosen s

hammer [34]

Answer:

There are $6\cdot 6=36$ different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

Step-by-step explanation:

If 4 screws must be chosen so that 2 have a Phillips head and 2 have a slotted head, then you have to choose 2 screws with a Phillips head from 4 screws with a Phillips head and 2 screws with a slotted head from 4 screws with a slotted head.

You can choose 2 screws with a Phillips head from 4 screws with a Phillips head in

$C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6$

different ways.

You can choose 2 screws with a slotted head from 4 screws with a slotted head in

$C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6$

different ways.

In total there are $6\cdot 6=36$ different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

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3 years ago