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4 years ago

I will make you the Brainliest

Mathematics

2 answers:

Stells [14]4 years ago

4 0

It’s the first one bc if u add -3 to the input it gives the outputs

mihalych1998 [28]4 years ago

3 0

Good morning ☕️

Answer:

<h2>Function Table number 1</h2>

Step-by-step explanation:

Input − 3 = output

-2 − 3 = -5

1 − 3 = -2

6 − 3 = 3

_______________________________

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Am I correct? The bottom answer is 4 if you can’t see it.

ivann1987 [24]

Answer:

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Step-by-step explanation:

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4 years ago

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3 years ago

find the equation of a cubic function whose graph passes through points (3,0) and (1,4) and is tangent to x-axis at the origin

Tomtit [17]

Answer:

y = -2x^2(x - 3)

Step-by-step explanation:

Preliminary Remark

If a cubic is tangent to the x axis at 0,0

Then the equation must be related to y = a*x^2(x - h)

(3,0)

If the cubic goes through the point (3,0), then the equation will become

0 = a*3^2(3 - h)

0 = 9a (3 - h)

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from which h = 3

From the second point, we get

4 = ax^2(x - 3)

4 = a(1)^2(1 - 3)

4 = a(-2)

a = 4 / - 2

a = -2

Answer

y = -2x^2(x - 3)

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3 years ago

- Polynomial Functions -For each function, state the vertex; whether the vertex is a maximum or minimum point; the equation of t

vladimir2022 [97]

EXPLANATION

Given the function f(x) = (x-6)^2 + 1

$\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$=x^2-12x+37$ $\mathrm{The\: parabola\: params\: are\colon}$ $a=1,\: b=-12,\: c=37$ $x_v=-\frac{b}{2a}$ $x_v=-\frac{\left(-12\right)}{2\cdot\:1}$ $\mathrm{Simplify}$ $x_v=6$ $y_v=6^2-12\cdot\: 6+37$

Simplify:

$y_v=1$

$\mathrm{Therefore\: the\: parabola\: vertex\: is}$ $\mleft(6,\: 1\mright)$ $\mathrm{If}\: a$ $\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}$ $a=1$ $\mathrm{Minimum}\mleft(6,\: 1\mright)$ $\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$y=x^2-12x+37$ $\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}$ $a=1,\: b=-12$ $x=\frac{-\left(-12\right)}{2\cdot\:1}$ $\mathrm{Refine}$

Axis of simmetry : x=6

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1 year ago

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tresset_1 [31]

Answer:

I believe the answer is function

Step-by-step explanation:

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4 years ago