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NikAS [45]
3 years ago
7

Determine whether each table represents a linear quadratic or exponential function

Mathematics
1 answer:
Kryger [21]3 years ago
7 0

Answer:

Table A: exponencial function

Table B: quadratic function

Table C: linear function

Step-by-step explanation:

Table A:

for x from 1 to 2, Y increased by 6

for x from 2 to 3, Y increased by 18

for x from 3 to 4, Y increased by 54

So we can see that the values of Y are being multiplied by 3 for each increment of x, so this table represents an exponencial function.

Table B:

for x from 1 to 2, Y increased by 6

for x from 2 to 3, Y increased by 10

for x from 3 to 4, Y increased by 14

The increment of Y is higher than a linear increment and smaller than an exponencial increment, so this is the quadratic function.

Table C:

for x from 3 to 6, Y increased by -2

for x from 6 to 9, Y increased by -2

for x from 9 to 12, Y increased by -2

The increment of Y is constant for the same increment of X, so this is a linear function.

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Length times width equals so your answer is 80

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3 years ago
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Find the area of the following figure below
abruzzese [7]

Answer:

30

Step-by-step explanation:

A=(a+b)/2xh

A=(6+9)/2x4

A=15/2x4

A=7.5x4

A=30

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2 years ago
Roots of Quadratics 50 PTS!!!
noname [10]

The values of k for the different quadratic equation solutions are as follows

a  the equation 2x² - x + 3k = 0 has two distinct real roots

  • k < 1/24

b. the equation 5x² - 2x + (2k − 1) = 0 has equal roots

  • k = 3/5

ci the equation -x² + 3x + (k + 1) = 0 has real roots

  • k > -3.25

d the equation 3kx² - 3x + 2 = 0 has no real solutions

  • k < ± 1.633

<h3>How to solve quadratic equations to get different answers</h3>

Quadratic equations of the form ax² + bx + c = 0 is solved using the formula

-b+\frac{\sqrt{b^{2}-4ac } }{2a}     OR     -b-\frac{\sqrt{b^{2}-4ac } }{2a}

The equation b² - 4ac is called the discriminant and it is used as follows

To solve the equation and get two real roots: 2x² - x + 3k = 0

  • b² - 4ac > 0

substituting the values gives

(-1)² - 4 * 2 * 3k > 0

1 - 24k > 0

1 > 24k

divide through by coefficient of k

k < 1/24

To solve the equation and get equal roots: 5x² - 2x + (2k − 1) = 0

  • b² - 4ac = 0

substituting the values gives

(-2)² - 4 * 5 * (2k - 1) = 0

4 - 40k + 20 = 0

-40k = -24

divide through by coefficient of k

k = 3/5

To solve the equation and get real roots  -x² + 3x + (k + 1) = 0

  • b² - 4ac > 0

substituting the values gives  

(3)² - 4 * -1 * (k+1) > 0

9 + 4k + 4> 0

4k > -13

divide through by coefficient of k

k > -3.25

To solve the equation and get  no real solutions  3kx² - 3x + 2 = 0

  • b² - 4ac < 0

substituting the values gives  

(-3)² - 4 * 3k * 2 < 0

9 - 24k² > 0

9 > 24k²

divide through by coefficient of k²

k² < 24/9

k < ± 1.633

Learn more about roots of quadratic equations: brainly.com/question/26926523

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4 0
8 months ago
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PLS HELP BEST ANSWER GETS BRAINLIEST
Blizzard [7]
Answer: 4 cakes
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4 0
2 years ago
Suppose a parabola has an axis of symmetry at x=-5, a maximum height of 9, and passes through the point (-7,1). Write the equati
Nutka1998 [239]

the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.

it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.

check the picture below.

so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)


\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9


\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill

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