. How many square feet of grass are there on a trapezoidal field with a height of 75 ft and bases of 125 ft and 81 ft?

sp2606 [1]

Answer:

(3 × 2x^8y^5) hope this helps

6 0

3 years ago

A gold mine is projected to produce $52,000 during its first year of operation, $50,000 the second year, $48,000 the third year,

marshall27 [118]

Answer:

its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

Step-by-step explanation:

Given that;

gold mine is projected to produce $52,000 during its first year

produce $50,000 the second year

produce $48,000 the third year

the mine is expected to produce for 20yrs; i.e n = 20

annual interest rate = 4% = 0.04%

now let P represent the present worth

we determine the present worth ;

Present worth ⇒ Cashflow(Uniform series present worth) - (2000)(uniform gradient present worth)

⇒Cashflow(P | A,i%,n) - (2000)(P | G,i%,n)

= 52000[ ((1+i)ⁿ - 1) / (i(1+i)ⁿ) ] - (2000)[ {((1+i)ⁿ - 1) / (i²(1+i)ⁿ))} - (n/i(1 + i)ⁿ) ]

= 52000[ ((1+0.04)²⁰ - 1) / (0.04(1+0.04)²⁰) ] - (2000)[ {((1+0.04)20 - 1) / ((0.04)²(1+0.04)²⁰))} - (20/0.04(1 + 0.04)²⁰) ]

= 52000[ ((1.04)²⁰ - 1) / (0.04(1.04)²⁰) ] - (2000)[ {((1.04)²⁰ - 1) / ((0.04)²(1.04)²⁰))} - (20/0.04(1.04)²⁰) ]

= 52000[ ((2.191123 - 1) / (0.04(2.191123) ] - (2000)[ {((2.191123 - 1) / ((0.0016)(2.191123))} - (20/0.04(2.191123) ]

= 52000(1.191123/0.08764) - (2000){( 1.191123/0.003506) - (20/0.87645)}

= 52000(13.59033) - (2000)(339.7582 - 228.1935)

= 52000(13.59033) - (2000)(111.5647)

= 706695.6 - 223129.4

= 483,566.2 ≈ 483,566

Therefore its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

4 0

3 years ago

You put $200 in a savings account. The account earns 2% simple interest

Black_prince [1.1K]

Answer:

interest total after 5 years is $20.82

Step-by-step explanation:

6 0

2 years ago

At 82°F, a certain insect chirps at a rate of 61 times per minute, and at 91°F, they chirp 124 times per minute. Write an equati

Ganezh [65]

Answer:

$y=7x-513$

Step-by-step explanation:

Let

x----> the temperature in degrees Fahrenheit

y ---->insect chirping rate in times per minute

we have the ordered pairs

(82,61) and (91,124)

step 1

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{124-61}{91-82}$

$m=\frac{63}{9}=7$

step 2

Find the equation of the line

we know that

The equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-intercept

we have

$m=7$ ---> the units are chirps per minute/degree F

take the point (82,61)

substitute and solve for b

$61=7(82)+b$

$61=574+b$

$b=-513$

substitute

$y=7x-513$

4 0

3 years ago

Can someone help me with this surface area plz plz thanks

Lunna [17]

Just multiply all of them and if you aren’t sure just search up how to find serfsce area

4 0

3 years ago