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True [87]
3 years ago
15

Use a graphing calculator to sketch the graph of the quadratic equation, and then state the domain and range. y=x^2+x+9 a. D: al

l real numbers R: (y ≤ 8.75) b. D: (x ≥ 0) R: (y ≥ 0) c. D: all real numbers R: (y ≥ 8.75) d. D: all real numbers R: (y ≥ -8.75)

Mathematics
2 answers:
bearhunter [10]3 years ago
8 0

Answer:

C

D: all real numbers R: (y ≥ 8.75)

Step-by-step explanation:

taurus [48]3 years ago
3 0

Answer: Option c. D: all real numbers R: (y ≥ 8.75)


Solution:

y=x^2+x+9

Domain: all real numbers = ( - Infinite, Infinite)

y=ax^2+bx+c

a=1 > 0, the parabola opens upward: Range: y ≥k

b=1

c=9

Vertex: V=(h,k)

h=-b/(2a)

h=-1/(2(1))

h=-1/2

h=-0.5

k=y=h^2+h+9

k=(-0.5)^2+(-0.5)+9

k=0.25-0.5+9

k=8.75

Range: y≥k

Range: y≥8.75




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Answer: 0.125%

Step-by-step explanation:

Let 1600 corresponds to the 100% value and 2 is a part of the total 100% value 1600.

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Answer:

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Answer:

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\textsf{Standard form}: \quad \sf 4x-y=-6

Step-by-step explanation:

<u>Given information</u>:

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\sf y-y_1=m(x-x_1)

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Substitute the given slope and point into the formula:

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<u>Slope-intercept form</u><u> of a linear equation</u>:

\sf y=mx+b

(where m is the slope and b is the y-intercept)

Substitute the given slope and point into the formula and solve for b:

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\implies \sf b=6

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\sf y=4x+6

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\sf Ax+By=C

(where A, B and C are constants and A must be positive)

Rearrange the found slope-intercept form of the equation into standard form:

\implies \sf y=4x+6

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\implies \sf 4x-y=-6

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