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Fudgin [204]
3 years ago
15

Find the distance between the points (3,-5) and (3,-2).

Mathematics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

3 units

Step-by-step explanation:

Distance between two points (x1,y1) and (x2,y2) on coordinate plane is given by

distance = \sqrt{(x1-x2)^2 +(y1-y2)^2  }

Given points in the problem are (3,-5)  and (3,-2).

Thus distance between points (3,-5)  and (3,-2).

distance = \sqrt{(x1-x2)^2 +(y1-y2)^2  }\\=>distance = \sqrt{(3-3)^2 +(-5-(-2))^2  }\\=>distance = \sqrt{(0)^2 +(-5+2)^2  }\\\\=>distance = \sqrt{0 +(-3)^2  }\\\\=>distance = \sqrt{9  }\\=>distance = 3

Distance between points (3,-5)  and (3,-2) is 3 units

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A rectangle has its base on the x axis and its upper two vertices on the parabola y = 9 − x2. (a) Draw a graph of this problem.
gtnhenbr [62]

Answer:

  • (a) see attached
  • (b) width: 2x; height: y = 9-x²
  • (c) A=2x(9-x²) . . . 0 ≤ x ≤ 3
  • (d) dA/dx = -6x² +18; x=±√3
  • (e) 12√3 units²

Step-by-step explanation:

(a) The attachment shows the graph of the parabola in blue. It also shows an inscribed rectangle in black.

(b) The upper right point of the rectangle is shown in the attachment as (x, y). The dimension y is the height of the rectangle. The x-dimension is half the width of the rectangle, which is symmetrical about the y-axis. Hence the width is 2x.

(c) As with any rectangle, the area is the product of length and width:

... A = (2x)(9 -x²) . . . . . the attachment shows a graph of this

... A = -2x³ +18x . . . . . expanded form suitable for differentiation

A suitable domain for A is where both x and A are non-negative: 0 ≤ x ≤ 3.

(d) The derivative of A with respect to x is ...

... A' = -6x² +18

This is defined everywhere, so the critical values will be where A' = 0.

... 0 = -6x² +18

... 3 = x² . . . . . . . divide by -6, add 3

... √3 = x . . . . . . . -√3 is also a solution, but is not in the domain of A

(e) The rectangle will have its largest area where x=√3. That area is ...

... A = 2x(9 -x²) = 2√3(9 -(√3)²) = 2√3(6)

... A = 12√3 . . . . square units . . . . ≈ 20.785 units²

7 0
3 years ago
I need graph help plz ​
Likurg_2 [28]

Answer:

Step-by-step explanation:

Part A

x-intercepts of the graph → x = 0, 6

Maximum value of the graph → f(x) = 120

Part B

Increasing in the interval → 0 ≤ x ≤ 3

Decreasing in the interval → 3 < x ≤ 6

As the price of goods increase in the interval [0, 3], profit increases.

But in the price interval of (3, 6] profit of the company decreases.

Part C

Average rate of change of a function 'f' in the interval of x = a and x = b is given by,

Average rate of change = \frac{f(b)-f(a)}{b-a}

Therefore, average rate of change of the function in the interval x = 1 and x = 3 will be,

Average rate of change = \frac{f(3)-f(1)}{3-1}

                                         = \frac{120-60}{3-1}

                                         = 30

6 0
3 years ago
Each row of the table represents a cone with a height of
kykrilka [37]
8171 9 inches and different radius
3 0
3 years ago
Using exact values (e.g. 10+4π), find the area of the following composite shape:
Angelina_Jolie [31]

Answer:

12 - (⅛)π cm²

Step-by-step explanation:

Rectangle + circle - semicircle

(4 × 3) + (pi × 1²) - ½(pi × 1.5²)

12 + pi(1 - 9/8)

12 - (⅛)π

3 0
3 years ago
Suppose y varies inversely with x, and y = 25 when x = 1/5. What is the value of y when x = 5 ?
34kurt

Answer:

y = 1

Step-by-step explanation:

Given that y varies inversely with x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition y = 25 when x = \frac{1}{5} , then

25 = \frac{k}{\frac{1}{5} } = 5k ( divide both sides by 5 )

5 = k

y = \frac{5}{x} ← equation of variation

When x = 5, then

y = \frac{5}{5} = 1

8 0
3 years ago
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