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belka [17]
3 years ago
9

A rectangle shed is fenced with 120ft long barbed wire. If length of the shed is twice as long as the width, that is the length?

Mathematics
1 answer:
creativ13 [48]3 years ago
7 0

Answer:

40 ft  

Step-by-step explanation:

<em>Data: </em>

(1) P = 120 ft

(2) l = 2w

<em>Calculations: </em>

(3)  P = 2l + 2w           Substitute (1) and (2) into (3)

  120 = 2(2w) + 2w     Multiply

  120 = 4w +2w           Combine like terms

  120 = 6w                   Divide by 6

(4) w = 20 ft                Substitute (4) into (2)

      l = 2 × 20

      l = 40 ft

<em>Check: </em>

120 = 2×40 + 2×20

120 = 80 + 40

120 = 120

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4 0
3 years ago
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F(x)=12x2-4x-8. G(x)=11x-6
AlekseyPX

Answer: Second option.

Step-by-step explanation:

Given the functions f(x) and g(x):

f(x)=12x^2-4x-8\\\\g(x)=11x-6

You need to divide them in order to find (\frac{f}{g})(x) asked in the exercise. Then:

(\frac{f}{g})(x) =\frac{12x^2-4x-8}{11x-6}

Now, it is important to remember that, by definition, the division by zero is not defined. Therefore, the denominator of the function cannot be zero.

Let's find the value of "x" for which the denominator of the function would be zero:

1. You need to make the denominator equal to zero:

11x-6=0

2. Finally, you must solve for "x":

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 (\frac{f}{g})(x) =\frac{12x^2-4x-8}{11x-6}, x=\frac{6}{11}

4 0
3 years ago
You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling
BARSIC [14]

Answer:

  (a) 15 m

  (b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

  d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

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(b) The <em>relative velocity</em> at 2.4 seconds is ...

  d'(t) = -5t

  d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

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We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

  d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

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<em>Comment on following distance</em>

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0
3 years ago
Which posulate proves the two triangles are congruent?
san4es73 [151]

Answer:

D ASA

Step-by-step explanation:

ASA

5 0
2 years ago
Answer these two question  ASAP
prohojiy [21]

Answer:

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6 0
3 years ago
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