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iren2701 [21]
4 years ago
7

Given a polynomial f(x), if (x - 1) is a factor, what else must be true?

Mathematics
2 answers:
Novosadov [1.4K]4 years ago
8 0
F(0) = 1  must be true
<span>If x=1,
 f(1)=(1-1)*g(1)=0*g(1)=0
 So, you get that f(1)=0 is true.
</span>hope this helps
kozerog [31]4 years ago
6 0

Answer:

f(1)=0

Step-by-step explanation:

we know that

If (x-1) is a factor of the polynomial f(x)

then

For x=1

the polynomial must be equal to zero

so

f(1)=0

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Answer:

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Step-by-step explanation:

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3 years ago
The distances male long jumpers for State College jump are approximately normal with a mean of 263 inches and a standard deviati
Effectus [21]

Answer:

His jump was of 272.45 inches

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 263, \sigma = 14

75th percentile

X when Z has a pvalue of 0.75. So X when Z = 0.675

Z = \frac{X - \mu}{\sigma}

0.675 = \frac{X - 263}{14}

X - 263 = 14*0.675

X = 272.45

His jump was of 272.45 inches

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3 years ago
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What is the simplest form of
azamat

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B

Step-by-step explanation:

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Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
Dafna1 [17]

Answer:

\huge\boxed{  \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} }  -  \frac{ {x}^{2} }{ {12}^{2} }  = 1}}}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • conic sections
  • PEMDAS
<h3>tips and formulas:</h3>
  • \sf hyperbola \:equation :  \\  \sf  \frac{ {x}^{2} }{ {a}^{2} }   -  \frac{ {y}^{2} }{ {b}^{2} }  = 1
  • vertices of hyperbola:(±a,0) and (0,±b) if reversed
  • \sf \: asymptotes :  \\ y =   \pm\frac{b}{a} x
<h3>given:</h3>
  • vertices: (0,±10)
  • the hyperbola equation is inversed since the vertices is (0,±10)
  • asymptotes:\pm \frac{5}{6}x
<h3>let's solve:</h3>
  • the asymptotes are in simplest and we know b is ±10

according to the question

  1. y =  \sf  \frac{5 \times 2}{6 \times 2} x \\ y =  \frac{10}{12} x

therefore we got

  • a=12
  • b=10

note: the equation will be inversed

let's create the equation:

  1. \sf substitute \: the \: value \: of \: a \: and \: b :  \\  \sf  \frac{ {y}^{2} }{ {10}^{2} }  -  \frac{ {x}^{2} }{ {12}^{2} }  = 1

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Two sides of a four sided figure have negative slopes which are the endpoints of the sides of this figure
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