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Ivahew [28]
3 years ago
6

Confused can someone help pls

Mathematics
1 answer:
Mariulka [41]3 years ago
8 0

For this case, we have that by definition, be given two functions f (x) and g (x). So:

(f-g) (x) = f (x) -g (x)

We have:f (x) = - 3x-5\\g (x) = 4x-2

So:

(f-g) (x) = - 3x-5- (4x-2)

We have to:

- * + = -\\(f-g) (x) = - 3x-5-4x + 2

Equal signs are added and the same sign is placed while different signs are subtracted and the sign of the greater is placed

(f-g) (x) = - 7x-3

Answer:

Option D

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Halim makes a profit of 120% on his cost price by selling a pair of shoes of $45. i) Find the cost pricer of the pair of shoes i
Rus_ich [418]

Answer:

a. $20.45

b. $24.55

c. $4.55

Step-by-step explanation:

In this question, we are asked to calculate the cost price and profit or loss selling a pair of shoes.

we proceed as follows

Firstly, we want to find the cost price given the selling price and the mark up percentage.

mathematically;

profit =( selling price - cost price)/cost price * 100%

here our profit is 120% with the cost price being $45. we plug these values

120 = 45-cp/cp * 100%

120cp = 100(45-cp)

120cp = 4500 -100cp

220cp = 4500

cp = 4500/220 = $20.45

His profit is selling price - cost price = 45 - 20.45 = $24.55

His profit selling same shoe for $25 is 25-20.45 = $4.55

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3 years ago
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Cos4theta+cos2theta/ cos4theta-cos2theta= _____
vovangra [49]

\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}

\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )

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