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Liono4ka [1.6K]
2 years ago
14

If the area of a park is exactly halfway between 2.4 and 2.5 acres, what is the area of the park

Mathematics
1 answer:
Ierofanga [76]2 years ago
6 0
Answer:

2.45

Explanation:

Halfway of 2.4 and 2.5 is 2.45 because .05 is the halfway mark based on the place value.
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The graph shows the distance in miles of a runner over x hours. What is the average rate of speed over the interval [9, 11]?
JulsSmile [24]
For this case what you should see is that for the interval [9, 11] the behavior of the function is almost linear.
 Therefore, we can find the average rate of change as follows:
 m = (y2-y1) / (x2-x1)
 m = (11-6) / (11-9)
 m = (5) / (2)
 m = 5/2
 Answer:
 the average rate of speed over the interval [9, 11] is: 
 D. 5 / 2
8 0
2 years ago
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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each equation with the operation yo
Tanzania [10]

The quadratic equations and their solutions are;

9 ± √33 /4 = 2x² - 9x + 6.

4 ± √6 /2 = 2x² - 8x + 5.

9 ± √89 /4 = 2x² - 9x - 1.

4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

2x² - 8x + 5, here a = 2, b = -8, c = 5.                                                  x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4.     24 can also be written as 4 × 6 and √4 = 2. So                                                                                     x = 8 ± 2√6 / 2×2= 4±√6/2.

2x² - 10x + 3, here a = 2, b = -10, c = 3.                                                   x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So                                                                              x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.

2x² - 8x - 3, here a = 2, b = -8, c = -3.                                                    x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4.     88 can also be written as 4 × 22 and √4 = 2. So                                                                             x = 8 ± 2√22 / 2×2 = 4± √22/2.

2x² - 9x - 1, here a = 2, b = -9, c = -1.                                                     x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4.                                          x = 9 ± √89 / 4.

2x² - 9x + 6, here a = 2, b = -9, c = 6.                                                    x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4.                                                                             x = 9 ± √33 / 4

To match we solve the monomials.

1. -15u^3 + 5u^3

Adding

-15u^3 + 5u^3=-10u^3

2.  10u^3 +(-5u^3)

Adding

10u^3-5u^3=5u^3

3. 10u^3 + 5u^3

Adding

10u^3 + 5u^3=15u^3

4.  5u^3+ (-10u^3)

Adding

5u^3-10u^3 =-5u^3

Two separate ways to find the answers.

7 0
2 years ago
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Marissa buys a licorice candy rope to share with some friends. The rope is 30 and two thirds inches long. She begins to cut the
adell [148]
9!
3.33 x 9 =(about)= 29.97
4 0
2 years ago
Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect
Colt1911 [192]

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid  

z = x2 + y2 when the coordinates (x,y,z)=(cos(πt), sin(πt), t) of the helix satisfy the equation of the paraboloid. That is, when

\bf (cos(\pi t), sin(\pi t), t)

But  

\bf cos^2(\pi t)+sin^2(\pi t)=1

so, the helix intersects the paraboloid when t=1. This is the point

(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The <em>tangent vector</em> to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface  

\bf z=x^2+y^2

at the point (-1, 0, 1) is given by

\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)

where

\bf f(x,y)=x^2+y^2

since

\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y

so, a normal vector to the tangent plane is

(-2,0,-1)

Hence, <em>a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane </em>is given by

\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.  

But we now

\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta

where  

\bf \theta= angle between the tangent vector and its projection onto the tangent plane. So

\bf \pi^2=(\sqrt{4+\pi^2}\sqrt{\pi^2+1})cos\theta\rightarrow cos\theta=\frac{\pi^2}{\sqrt{4+\pi^2}\sqrt{\pi^2+1}}=0.8038

and

\bf \theta=arccos(0.8038)=0.6371\;radians

7 0
1 year ago
Chris built a rectangular snow fort with a perimeter of 24 feet. The length of the fort was 8 feet less than 3 time the width. W
Ira Lisetskai [31]
24= x+y
X= 3y-8
24= 3y-8+y
24=4y-8
32=4y
Y=8
X=16
4 0
1 year ago
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