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3 years ago

Which expression has the same sum as the one below?

Mathematics

2 answers:

BartSMP [9]3 years ago

7 0

Answer:

6+(-15)

Step-by-step explanation:

+ multiplied by minus is minus so that is 6-15

docker41 [41]3 years ago

6 0

Step-by-step explanation:

The First one and the last one

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Choose the correct simplification of the expression (d^3)^5

marin [14]

(d^3)^5 is equal to d^(3^5), which is equal to d^243.

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Leiff goes online to buy a new video game. He finds a site that currently has a promotion of 15% off on all orders over $50. Lei

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If the original price tag of the video game is $128, to get its discounted price at 15%, that means what Leiff paid for was (100-15) = 85% of the original price. So, the price of the video game then becomes (0.85 x 128) = $108.80. Since Leiff has to pay an additional 5.3% of the new price, that's (100 + 5.3) = 105.3% of the new price which is (1.053 x 108.8) = $114.57 Adding the shipping fee for the videogame, 114.57 + 4.75 = $119.32 Thus, the total price is <span>$119.32</span>

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lilavasa [31]

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Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (

IrinaVladis [17]

Answer:

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14 and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = $\dfrac{9+9}{2}$

median = $\dfrac{18}{2}$

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = $\dfrac{12+13}{2}$

Median = $\dfrac{25}{2}$

Median = 12.5

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = Q3 - Q1

The interquartile range = 12.5 - 9

The interquartile range = 3.5

7 0

3 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

5 0

3 years ago