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2 years ago

What is the simplified form of (-3xy)(5xy-1)?

Mathematics

1 answer:

harkovskaia [24]2 years ago

4 0

Answer:

-15x²y²+3xy

Step-by-step explanation:

((−3x)(y))(5xy−1)

=((−3x)(y))(5xy+−1)

=((−3x)(y))(5xy)+((−3x)(y))(−1)

=−15x²y²+3xy

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A shopkeeper buys pens at the rate of ₹ 75 per 100. for how much should he sell each pen so as to make a gain of 15%? please hel

Lady bird [3.3K]

Answer:

He should sell each pen of Rs.0.8625 so as to make a gain of 15%

Step-by-step explanation:

Cost price of 100 pens = Rs.75

Cost price of 1 pen= $\frac{75}{100}$ = 0.75

Gain%= 15%

Gain= $15%$ $X 0.75$ $= 0.1125$

SP = CP+Gain = 0.75+0.1125=0.8625

<u>So, He should sell each pen of Rs.0.8625 so as to make a gain of 15%</u>

#Learn more:

By selling 60 pens a shopkeeper gain the cp of 15 pen. If he bought 20 pens in rs36. Find the sp of one pen

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https://brainly.in/question/7469007

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3 years ago

Determine if its possible to form a triangle using the set of segments with the given measurement. explain please! 12 m, 5 m, 18

IRINA_888 [86]

Adding together the shorter two sides results in 17 m. If these two sides were laid end to end, the sum of their sides (17 m) would be less than the length of the third side. Thus, NO triangle could be constructed with this set of line segments.

6 0

3 years ago

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

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4 0

2 years ago

Am I right about this or..?

Arisa [49]

Answer:

Correct, as 4 is the only number repeating twice

4 0

3 years ago

The price of all rail season tickets to London increased by 4%. The price of a rail season ticket from Cambridge to London incre

kirill115 [55]

Percentage increase in the rate of all rail season tickets to London = 4%
Price of the rail ticket from Cambridge to London increased by = 121.60
Let us assume the price before the increase in the fare = x
Then
4% * x = 121.60 dollars
(4/100) * x = 121.60 dollars
x/25 = 121.60 dollars
x = 25 * 121.60 dollars
= 3040 dollars
So the price of the ticket from Cambridge to London before the increase in price was 3040. I hope the procedure is simple enough for you to understand.

8 0

3 years ago