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Artyom0805 [142]
3 years ago
14

What is the range of this function (I NEED AN ANSWER RLLY FAST PLSSSS)

Mathematics
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

d

Step-by-step explanation:

the first oval's domain and the second's range. so you just choose the choice that includes all the values in the ranger.

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On the same coordinate plane mark all points (x,y) such that
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It is A because it can’t be B and it can’t be C so my answer is A
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3 years ago
How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}

Therefore, the directrix is;

y=-\frac{3}{2}

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Answer:

I BELIEVE ITS C

Step-by-step explanation:

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Evaluate the expression below x/2 +8y-15 for x = -12 and y=3
jeka94
<span>(x/2)+8y-15 = (-12/2)+8*3-15 = -6+24-15 = 33</span>
3 0
3 years ago
What are the AGI and taxable income?
andreev551 [17]

Answer:

1. Find the AGI and taxable income: Gross Income $30,856 Adjustments $750 1 Exemption $8200 Deduction $2,300

AGI: $31,200 and $20,601 $30106 --- ANSWER: 30,106 (30,856-750)

Taxable Income: $19,606 $29,586 and $18,505 $28,863 and $17,636 1 points--- ANSWER 19,606

2. QUESTION 5 Find the AGI and taxable income. Gross Income $67,890

Adjustments $0 3 Exemptions $24,600 Deduction $1469

AGI: $69,440 and $45,300 $68,990 and $42,831 $67,890 --- ANSWER:

67,890

Taxable Income: $41,821 $65,551 and $44,821 1 points --- ANSWER: 41,821 (67,890-24,600-1,469)

3. QUESTION 6 Find the AGI and taxable income. Gross income $19,723 Adjustments $255 1 Exemption $8200 Deduction $1430 $19,4

AGI: 19,468 (19,723-255)

Taxable Income: 9,838 (19,468-8,200-1,430)

Step-by-step explanation:

Still stuck? Get 1-on-1 help from an expert tutor now.

3 0
2 years ago
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