Step-by-step explanation:
4x + 7x + 2° = 90° { being complementary angles }
11x = 90° - 2°
11x = 88°
x = 88° / 11
x = 8°
<YVZ = <em>7</em><em> </em><em>*</em><em> </em><em>8</em><em>°</em><em> </em><em>+</em><em> </em><em>2</em><em>°</em><em> </em><em>=</em><em> </em><em>5</em><em>8</em><em>°</em>
<em>Hope </em><em>it </em><em>will </em><em>help </em><em>:</em><em>)</em>
Answer:
$507.00.
Step-by-step explanation:
First, converting R percent to r a decimal
r = R/100 = 6.5%/100 = 0.065 per year,
then, solving our equation
I = 1300 × 0.065 × 6 = 507
I = $ 507.00
The simple interest accumulated
on a principal of $ 1,300.00
at a rate of 6.5% per year
for 6 years is $ 507.00.
Using the Sine Rule:-
20 /sin 45 = BC / sin 30
BC = 20 sin 30 / sin45
= 20 * 1/2 / 1 / sqrt2
= 20 * 1/2 * sqrt2
= 10 sqrt2 answer
Answer:
ai) 73000 gal/yr
aii) $730 per year
b) 1000 days
Step-by-step explanation:
ai) The water usage per day is ...
(4 showers/day)(10 min/shower)(5 gal/min) = 200 gal/day
Then the usage per year is ...
(200 gal/day)(365 days/yr) = 73,000 gal/yr
__
aii) The cost of electricity is the cost of heating half the water usage, so is ...
(0.20 kWh/gal)(1/2)(73,000 gal/yr)($0.10/kWh) = $730/yr
__
b) The current daily cost of electricity for heating water is ...
($730/yr)/(365 days/yr) = $2/day
If the cost is cut in half, it will be $1 per day. That means the savings is $1 per day, so it will take 1000 days to recover the $1000 initial cost. (Effectively, the average cost for the first 1000 days is the same as if the water heater had not been replaced.)
