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stiks02 [169]
3 years ago
15

Please help me with fuction problems.

Mathematics
2 answers:
DerKrebs [107]3 years ago
8 0

1) For each of these, keep in mind vertex form: f(x)=a(x-h)^2+k. With vertex form, a is the direction and width, h is the horizontal placement of the vertex, and k is the vertical placement. For the first one, notice that "a" is positive 1, so it faces up. This means that D, the one facing down, cannot be the answer. "h" is 1, so we will move the vertex to the right one unit (keep in mind (x-h), so if it were to be (h+3) you would move it to the left, not the right). "k" is -3, so we would move the vertex down 3 units. That said, the vertex should be at (1,-3) so the answer is C, or the one right below the first one.

2) The graph of f(x)=|2x| translated 5 units to the left means that h is equal to -5. When we plug -5 into vertex form, it should look like: g(x)=|2(x+5)|. The answer to this is A.

3) The equation for reflection on the x axis is f(x)=-a(x-h)+k. So, if the parent function f(x)=4|x| were to be reflected on the x axis, the function would look like this: g(x)=-4|x|. The answer to this should be B.

4) Since h=1 and k=0 in the function f(x)=-3|x-1|, the vertex will be (1,0).

5) This can also be written as g(x)=|x|-3. This means that k=-3, and will be a vertical translation of 3 units down.

mezya [45]3 years ago
4 0
The answer will be C
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Answer:

D. All real numbers

Step-by-step explanation:

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The domain [y] is limited, but the range [x] is best described as all real numbers

(if it's hard to remember which is which, d comes before r alphabetically, and x comes before y alphabetically)

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2 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
What is 2/3 times 26 but estimated ?
Elena-2011 [213]
It would be about 17, I think
3 0
3 years ago
Read 2 more answers
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