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yKpoI14uk [10]
2 years ago
9

What is the final amount if 392 is increased by 9% followed by a 3% decrease?

Mathematics
1 answer:
strojnjashka [21]2 years ago
5 0

Answer:

392/100 X 109 = 427.28

427.28/100 X 97 = 414.4616

414.46

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During a clothing store's Bargin Days, the regular price for T-shirts is discounted to $8.25. You have an additional coupon for
kakasveta [241]

Answer:

p(t) = 8.25t - 5

b) The cost of 11 T-shirts during bargain days is $85.75

Step-by-step explanation:

We are given the following information in the question:

Discounted cost of one T-shirt = $8.25

Additional off on coupon = $5.00

Let t be the number of T-shirts bought during the bargain days.

a) Then, the price of t T-shirts with the discount applied will be given by the function:

p(t) = \text{Number of T-shirts}\times \text{Discounted price for 1 T-shirt} - \text{Coupon Off}\\p(t) = 8.25t - 5

b) Price during Bargain Days for 11 shirts

We put t = 11 in the above function:

p(t) = 8.25t - 5\\p(11) = 8.25(11) - 5 = 85.75

Thus, the cost of 11 T-shirts during bargain days is $85.75

4 0
3 years ago
What is the solution to the inequality? -2/3(2x - 1/2) ≤ 1/5x - 1 Express your answer in interval notation.
almond37 [142]

ANSWER


-\frac{2}{3}(2x-\frac{1}{2})\le \frac{1}{5}x-1


Multiply through by LCM of 15


(15) \times -\frac{2}{3}(2x-\frac{1}{2})\le 15(\frac{1}{5}x-1)





-10(2x-\frac{1}{2})\le 3x-15



Expand brackets to obtain,



-20x+5\le 3x-15



Group like terms



15+5\le 20x+3x




20\le 23x


\frac{20}{23}\le x




x\ge \frac{20}{23}


In interval form it is written as


[\frac{20}{23}, \infty)







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Answer:

0.2c-c

Step-by-step explanation:

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andrezito [222]

Answer:

if there were 8 cats then theres 8 cats, or 32 dogs

Step-by-step explanation:

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1 year ago
which function grows at fastest rate for increasing values of x? h(x) = 6x^2+1 , f(x) = 9x +14, g(x)= 4^x
mezya [45]
Hi there! The answer is g(x) = 4^x.

The function g(x) is an exponential function, and these functions grow faster than linear or quadratic formulas.
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