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Rudik [331]
3 years ago
8

#7 please I don’t get how to do this

Mathematics
1 answer:
motikmotik3 years ago
4 0

Get A Better PICTURE

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Please help!
crimeas [40]

Answer:

The zeros are:

x =4, x=2, x = 5

  • The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

0=\left(x-4\right)^2\left(x^2-7x+10\right)

switch sides

\left(x-4\right)^2\left(x^2-7x+10\right)=0

as

x^2-7x+\:10=\left(x-2\right)\left(x-5\right)

so

\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0

Using the zero factor principle

  • \mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x =4, x=2, x = 5

Thus, the zeros are:

x =4, x=2, x = 5

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

  • The function has three distinct real zeros.

Hence, option (B) is true.

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2 years ago
There are 4 jacks and 13 clubs in a standard, 52-card deck of playing cards. What is the probability that a card picked at rando
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Answer:

16/52, or 4/13.

Step-by-step explanation:

First, since we know that the question is asking for the probability of a club <u>or</u> a jack, we know that we have to add the two probabilities. The first probability is that of picking a club, which is 13/52. The probability of picking a jack (be sure not to overlap; don't double count the jack of clubs) is 3/52. Adding these two gives us 13/52+3/52=16/52, which simplifies to 4/13.

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Answer:

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Step-by-step explanation:

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