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blondinia [14]
3 years ago
8

Mr Leonard wants to top-up his central heating oil tank. The oil tank can take up to 1200 litres of oil. There are already 450 l

itres of oil in the tank. The price of oil is 81.5p per litre. As a loyal customer, Mr Leonard gets a 7.5% discount on the price of the oil. How much discount does Mr Leonard get on this purchase? Give your answer in £
Mathematics
1 answer:
garik1379 [7]3 years ago
7 0

Mr. Leonard gets £ 45.84375 as discount

<em><u>Solution:</u></em>

The oil tank can take up to 1200 liters of oil

There are already 450 liters of oil in the tank

The remaining oil which is to be added is given by :

Remaining oil = 1200 - 450 = 750 liters

The price of oil is 81.5 p per liter

<em><u>Then calculate the total price of 750 L of oil:</u></em>

Total\ Price = 750 \times 81.5 = 61125

Thus total price is 61125 p

Mr Leonard gets a 7.5% discount on the price of the oil

Therefore,

Discount amount = 7.5 % of 61125

Discount\ amount = 7.5 \% \times 61125\\\\Discount\ amount = \frac{7.5}{100} \times 61125\\\\Discount\ amount = 0.075 \times 61125\\\\Discount\ amount = 4584.375

Thus he gets 4584.375 p

We convert p to £

1 p = 0.01£

4584.375 x 0.01 = £ 45.84375

Thus he gets £ 45.84375 as discount

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5n = 10^2
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3 years ago
What is the slope of the line that passes through the points (1,-5) and (3,5)?
Lana71 [14]

Answer:

a) 5

Step-by-step explanation:

Put them into slope intercept form AKA y=mx+b: -5=m1+b <em>and</em> 5=m3+b

Subtract the equations to get -10=-2m

Divide both sides by two to get: 5=m

Hope this helps! (^-^)

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3 years ago
HELP Use either law of sines or law of cosine. Need help on this problem! show work please!​
Marina86 [1]

Answer: x = 15.035677095729 approximately

Round this however you need to.

=================================================

Explanation:

I'm assuming you want to find the value of x, which your diagram is showing to be the length of segment QR.

If so, then we'll need to find the measure of angle Q first. Using the law of sines, we get the following:

sin(Q)/q = sin(R)/r

sin(Q)/PR = sin(R)/PQ

sin(Q)/13 = sin(85)/19

sin(Q) = 13*sin(85)/19

sin(Q) = 0.6816068987

Q = arcsin(0.6816068987) ... or ... Q = 180-arcsin(0.6816068987)

Q = 42.9693397461 ... or ... Q = 137.0306602539

These values are approximate.

----------------

Now if Q = 42.9693397461 approximately, then angle P is

P = 180-Q-R

P = 180-42.9693397461-85

P = 52.0306602539

Similarly, if Q = 137.0306602539 approximately, then,

P = 180-Q-R

P = 180-137.0306602539-85

P = -42.0306602539

A negative angle is not possible, so we'll ignore Q = 137.0306602539

----------------

The only possible value of angle P is approximately P = 52.0306602539

Let's apply the law of sines again to find side p, aka segment QR

sin(P)/p = sin(R)/r

sin(P)/QR = sin(R)/PQ

sin(52.0306602539)/x = sin(85)/19

19*sin(52.0306602539) = x*sin(85)

19*sin(52.0306602539)/sin(85) = x

x = 15.035677095729

This value is approximate.

Round this value however you need to.

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