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Anuta_ua [19.1K]
10 months ago
5

Solve for x 73 =(9/10x)+19

Mathematics
1 answer:
Yuliya22 [10]10 months ago
3 0
73=(\frac{9}{10}x)+19

To solve x:

1. Substract 19 in both sides of the equation:

\begin{gathered} 73-19=(\frac{9}{10}x)+19-19 \\  \\ 54=\frac{9}{10}x \end{gathered}

2. Multiply both sides of the ewuation by 10/9:

\begin{gathered} 54\cdot\frac{10}{9}=\frac{9}{10}x\cdot\frac{10}{9} \\  \\ \frac{540}{9}=x \\  \\ 60=x \end{gathered}<h2>Then, the value of x is 60 (x=60)</h2>
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What are the areas of each smaller rectangle and the large rectangle?
zhannawk [14.2K]

Answer:

Area of part 1: 3 * 5\sqrt{2} = 15\sqrt{2}

Area of part 2: 3\sqrt{2} * 5\sqrt{2}=30

Area of entire rectangle: 30 + 15\sqrt{2}

Step-by-step explanation:

The area of a rectangle can be found by using formula : length * width.

The area of part 1 can be simplified to this:

3 * 5\sqrt{2} = 15\sqrt{2}

Area of part 2:

3\sqrt{2} * 5\sqrt{2}=30

The area of the entire rectangle can simply be added from part 1 and part 2.

For area 2, remember multiplying the square root of 2 twice nullifies the square root. So it becomes 15 * 2, which is equal to 30.

7 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

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2 years ago
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3 years ago
Five workers are hired to seed a field by hand. Each is given a plot which is 6x12 feet in size. What is the total area of the f
Lorico [155]

Answer:

360 feet

Step-by-step explanation:

multiple 6 by 12 to get the area of the individual plot of land given to a single worker.

Then multiply by 5 for all 5 workers getting the same plot of land.

3 0
3 years ago
A plant measures 0.5 in. at the end of Week 1 and 14 in. at the end of Week 5. find the rate of change
CaHeK987 [17]
The change in growth was 14 - 0.5 = 13.5 inches.
Since it is over the course of 5 weeks, divide 13.5 by 5.
13.5/5 = 2.7
2.7 = 270%

The rate of change was 270%. Hope this helps!
8 0
3 years ago
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