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3 years ago

Which function is a quadratic function? c(x) = 6x + 3x3 d(x) = x – 8x4 p(x) = –5x – x2 k(x) = 2x2 + 9x4

Mathematics

2 answers:

jonny [76]3 years ago

8 0

Answer:

p(x) = –5x – x2

Step-by-step explanation:

just took the test

Citrus2011 [14]3 years ago

7 0

Easy
quadratic means 2nd degree

means highest power of exponent is 2
so like x^2

first one
we have 3x^3, that is 3rd degree,nope

2nd one
we have 8x^4, nope, that is 4th degree

3rd one
we have -x^2, yep

4th one, we have 9x^4, that is 4th degree

answer is 3rd one or p(x)=-5x-x^2

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Algebra 1. i just need help with understanding multiple step equations like...

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3 years ago

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Find the perimeter of a quadrilateral with vertices at C (-1, 2), D (-2, -1), E (2, -2), and F (1, 1). Round your answer to the

babymother [125]

Answer:

B 12.68

Step-by-step explanation:

you will need to find the distance between each segment-

so the length of DC, CF, FE, ED

$d=\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1}) ^{2} }$

DC:

D(-2,-1) C(-1,2)

$d=\sqrt{(-1--2)^{2} +(2--1)^{2} } \\d=\sqrt{1^{2}+3^{2} } \\d=\sqrt{10} =3.16$

CF:

C(-1,2) F(1,1)

$d=\sqrt{(1--1)^{2} +(1-2)^{2} } \\d=\sqrt{2^{2}+-1^{2} } \\d=\sqrt{4+1} \\d=\sqrt{x} 5=2.24$

F(1,1) E(2,-2)

$d=\sqrt{(2-1)^{2} +(-2-1)^{2} } \\d=\sqrt{1^{2} +(-3)^{2} } \\d=\sqrt{1+9} \\d=\sqrt{10}= 3.16$

E(2,-2) D(-2,-1)

$d=\sqrt{(-2-2)^{2} +(-1--2)^{2} } \\d=\sqrt{(-4)^{2}+1^{2} } \\d=\sqrt{16+1} \\d=\sqrt{17}= 4.12\\$

3.16+2.24+3.16+4.12= 12.68

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3 years ago

N = 32 ÷ (5 - 1)<br> N = 32 ÷ ?<br> N =?

Roman55 [17]

Answer

N = 32 ÷ (5 - 1)

N = 32 ÷ 4

N =8

Step-by-step explanation:

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3 years ago

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5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

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3 years ago

Which expression has a value that is equivalent to 5n + 13 when n=−5 ?

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