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3 years ago

15

Please some help for question 1

2 answers:

o-na [289]3 years ago

7 0

Answer:

f (3) is equal to 31

Step-by-step explanation:

f (3) means that x is equal to 3 [hence f (x) ---> f (3)]

Plugin 3 into all the x values to solve:

4(3)^2 - 2(3) + 1

4(9) - 6 + 1

36 - 6 + 1

30 + 1

31

f (3) is equal to 31

alexgriva [62]3 years ago

7 0

Answer:

31

The answer to this problem is 31.

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Solve for d:<br> 2d+1/6=6(1/3d+1)

taurus [48]

2d+1/6=2d+6 (I distributed the 6 to what was in the parentheses)
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koban [17]

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Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Hich function has a range of {y|y ≤ 5}?

marshall27 [118]

Answer:

$f(x)=-(x-4)^{2}+5$

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

if a>0 -----> the parabola open upward (vertex is a minimum)

if a<0 -----> the parabola open downward (vertex is a maximum)

<u>Verify each case</u>

case A) $f(x)=(x-4)^{2}+5$

The vertex is the point $(4,5)$

$a=1$

a>0 -----> the parabola open upward (vertex is a minimum)

The range is the interval--------> [5,∞)

$y\geq5$

case B) $f(x)=-(x-4)^{2}+5$

The vertex is the point $(4,5)$

$a=-1$

a<0 -----> the parabola open downward (vertex is a maximum)

The range is the interval--------> (-∞,5]

$y\leq 5$

case C) $f(x)=(x-5)^{2}+4$

The vertex is the point $(5,4)$

$a=1$

a>0 -----> the parabola open upward (vertex is a minimum)

The range is the interval--------> [4,∞)

$y\geq4$

case D) $f(x)=-(x-5)^{2}+4$

The vertex is the point $(5,4)$

$a=-1$

a<0 -----> the parabola open downward (vertex is a maximum)

The range is the interval--------> (-∞,4]

$y\leq 4$

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3 years ago

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jeyben [28]

Answer:

m=3

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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4 years ago