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vesna_86 [32]
4 years ago
10

The perimeter of a square is represented by the expression 32x - 12.8

Mathematics
1 answer:
kow [346]4 years ago
8 0

Answer:

4(8x-3.2)

Step-by-step explanation:

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a rectangle plot measure 20ft. by 30ft. A 3-ft.-wide sidewalk surrounds it.Find the area of the sidewalk.
goldenfox [79]
Area of a rectangle = length (l) * width (w)
A = 30ft * 20ft
A = 600 sq ft

Now the width of a sidewalk that surroundeds it = 3 ft
so now the area of the rectangle with sidewalk= 30+3ft * 20+3ft
A = (33*23) ft
A = 759 sg ft

Area of the sidewalk = 759 - 600
A = 159 sq ft
3 0
3 years ago
The first three terms of a sequence are given. Round to the nearest thousandth (if
tangare [24]

The 7th term of the sequence to the nearest thousandth is 223.949

<h3>What are geometric sequences?</h3>

These are sequence that increases in an exponential form.

The formula for calculating the nth term of a geometric sequence is expressed as:

Tn = ar^n-1

Given the following parameters

a = 75

r = 1.2

n = 7

Substitute the given parameter into the formula

T7 = 75(1.2)^6
T7 = 223.949

Hence the 7th term of the sequence to the nearest thousandth is 223.949

Learn more on geometric sequence here: brainly.com/question/9300199

3 0
2 years ago
Which is the graph of f(x) = x2 - 2X + 3?
Svet_ta [14]
The graph is the one with coordinates (1,2) as the center.

8 0
3 years ago
Read 2 more answers
What is the solution to the system of equations below? 2 x + 3 y = 17. 3 x + 6 y = 30. (Negative 7, 24) (3, 4) (4, 3) (24, Negat
butalik [34]

Answer:

work is shown and pictured

6 0
3 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
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