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notsponge [240]
3 years ago
13

The perimeter of a rectangular garden is 160 m and its width is 30 m. Find the cost of leveling it at the rate of $2.50 per sq.m

. $5,462 $3,750 $3,840 $9,840
Mathematics
1 answer:
nalin [4]3 years ago
5 0

Answer:

$12,000.

Step-by-step explanation

Length of garden = 160 m

Width of garden = 30 m

area of rectangle = length * width

substituting vale of length and width for rectangular garden we have

area of rectangular garden = 160 m * 30 M = 4800 sq. m

cost of leveling one sq. m is $2.50

and it is required to level 4800 sq. m

hence to total cost to level rectangular garden = area of rectangular garden * cost of leveling one sq. m of space

=4800 * 2.5

= 12,000

Therefore cost of leveling rectangular garden is $12,000.

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Answer:

x = 3

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Step-by-step explanation:

Using theorem of similar triangles, we have,

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7(6 + x) = 10.5(6)

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Thus:

7.5/y = (7 + 3.5)/7

7.5/y = 10.5/7

Cross multiply

7.5*7 = 10.5*y

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How to write 3/8 two other ways
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3 years ago
Read 2 more answers
If 25% of a number is 65 and 60% of the same number is 156, find 35% of that number.
Goshia [24]

Answer:

35% of that number is 91.

Step-by-step explanation:

we need to first find 25% of what number equals 65.

65×100÷25 = 260

Then, if you use the same method again, 60% of 260 would be 156.

We know "that" number is 260, and all we need to do is find 35% of it.

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8 0
3 years ago
Determine the current through each of the LEDs in the circuits below. Which LED will be
Anika [276]

a. I = 6. 1 × 10^-4 A

b. I = 2. 6 × 10^-3 A

c. I = 0. 04 A

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

<h3>How to determine the current</h3>

The formula for finding current

I = V/R

Where v = voltage

R = resistance

A. V = 12V

R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series

I = \frac{12}{19700}

I = 6. 1 × 10^-4 A

B. V = 9V

R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series

I = \frac{12}{4700}

I = 2. 6 × 10^-3 A

C.  V=  12V

1/R = \frac{1}{750} + \frac{1}{1200} + \frac{1}{950 } = 3. 22 × 10^-3

R = \frac{1}{3. 22 * 10 ^-3} = 310. 56 Ω

I = \frac{12}{310. 56}

I = 0. 04 A

It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

Learn more about Ohms law here:

brainly.com/question/14296509

#SPJ1

8 0
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