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mojhsa [17]
2 years ago
9

Gary used landscape timbers to create a border around a garden shaped like a right triangle. The longest two timbers he used are

12 feet and 15 feet long. Which is closest to the length, in feet, of the shortest timber?
Mathematics
1 answer:
Cloud [144]2 years ago
8 0

Answer:

9 feet

Step-by-step explanation:

Given:

The border of the garden is a right angled triangle.

Two lengths are given as 12 ft and 15 ft.

Let the length of the shortest timber be 'x' feet.

Now, in a right angled triangle, the longest length is called the hypotenuse.

As 15 feet is the largest length, it is the hypotenuse of the triangle. Now, applying Pythagoras theorem, we get:

(Leg1)^2+(Leg2)^2=(Hypotenuse)^2\\x^2+12^2=15^2\\x^2+144=225\\x^2=225-144\\x^2=81\\x=\pm \sqrt{81}=\pm 9

The negative value is neglected as length can never be negative.

Therefore, the length of the shortest timber is 9 feet.

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On a coordinate plane, square P Q R S is shown. Point P is at (4, 2), point Q is at (8, 5), point R is at (5, 9), and point S is
sergey [27]

Answer:

(D)The midpoint of both diagonals is (4 and one-half, 5 and one-half), the slope of RP is 7, and the slope of SQ is Negative one-sevenths.

Step-by-step explanation:

  • Point P is at (4, 2),
  • Point Q is at (8, 5),
  • Point R is at (5, 9), and
  • Point S is at (1, 6)

Midpoint of SQ =\frac{1}{2}(1+8,5+6)=(4.5,5.5)

Midpoint of PR =\frac{1}{2}(4+5,2+9)=(4.5,5.5)

Now, we have established that the midpoints (point of bisection) are at the same point.

Two lines are perpendicular if the slope of one is the negative reciprocal of the other.

In option D

  • Slope of RP =7
  • Slope of SQ  =-\dfrac17

Therefore, lines RP and SQ are perpendicular.

Option D is the correct option.

6 0
2 years ago
The chart indicates the time, speed, and velocity of five runners.
Nuetrik [128]
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2 years ago
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Answer:

$12.50

Step-by-step explanation:

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4 0
2 years ago
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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
MAVERICK [17]
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
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Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
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2 years ago
sandy is observing the velocity of a runner at different times. After one hour, the velocity of the runner is 4 km/h. After two
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A) Let x stand for time, y stand for velocity.
We are given the points (2,50), (6, 54). We can make a line using the slope intercept form
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plug in point (2,50) to find b
50 = 1(2) + b
50-2 = b
48 = b
the equation is  y = 1x + 48
Make standard form.
<span>x - y = -48</span>
7 0
2 years ago
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