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Gennadij [26K]
3 years ago
5

Could someone please help me :) its easy I think - not to me though...

Mathematics
1 answer:
lara [203]3 years ago
3 0
Lets find the unit rate of both of the shirts.

5 for 63 

4 for 54 

63/5 = 12.60

54/4 = 13.50

Now we have our unit rates!

Greg =   $12.60 per shirt
Zachary =  $13.50 per shirt

Greg's shirt were cheaper than his brothers, so it has a better cost. 
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Need help with geometric sequences <br> 30 points
NeTakaya

Answer:

a_{n} = 6(5)^{n-1}

Step-by-step explanation:

The n th term of a geometric sequence is

a_{n} = a(r)^{n-1}

where a is the first term and r the common ratio

Here a = 6 and r = 30 ÷ 6 = 5, thus

a_{n} = 6(5)^{n-1}

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In the equation c*2=22 what is the next step in the equation solving sequence?
romanna [79]
Divide 22 by 2 to get "c" isolated.
3 0
3 years ago
Verify the identity algebraically:<br> Csc(-x)tanx =-secx
weqwewe [10]

Step-by-step explanation:

Recall that

\sin(-x) = -\sin x

Therefore,

\csc(-x) = \dfrac{1}{\sin(-x)} = -\dfrac{1}{\sin x}

so

\csc(-x)\tan x = \left(-\dfrac{1}{\sin x}\right)\left(\dfrac{\sin x}{\cos x}\right)

\:\:\:\:\:\:\:\:\:= -\dfrac{1}{\cos x} = -\sec x

5 0
2 years ago
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
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