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4 years ago

Pls helpp!! i need the answers NOW is this correct??

Mathematics

1 answer:

julsineya [31]4 years ago

6 0

Yeah it isssssssssssssss

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What is the answer to these problems

BaLLatris [955]

Its 35543562454645656466 and 3568899687698797997978 also 5679697-5679567-79567 two more 8439485949834595893458485 abd 84853945934853495834593458

4 0

4 years ago

If f(x) = 4x – 3 and g(x) = 8x + 2, find each function value

vladimir1956 [14]

Answer:

f[g(3)] = 100

g[f(5)] = 138

g{f[g(-4)]} = -982

Step-by-step explanation:

f(x) = 4x - 3

g(x) = 8x + 2

a. f[g(3)]

First find g(3) by putting x = 3 in g(x) function

g(3) = 8(3) + 2 = 24 +2 = 26

now put this g(3) = 26 as x in f(x)

f(g(3)) = f(26) = 4(26) - 4 = 104 - 4 = 100

b. g[f(5)]

First lets find f(5)

f(x) = 4x - 3

put x = 5 above

f(5) = 4(5) - 3 = 20 - 3 = 17

put this f(5) = 17 as x in g(x)

g(f(5)) = g(17) = 8(17) + 2 = 138

c.g{f[g(-4)]}

First lets solve the inner most function

g(-4) = 8(-4) +2 = -30

put g(-4) = -30 in f(x) to find f(g(-4))

f(g(-4)) = f(-30) = 4(-30) - 3 = -123

put f(g(-4)) = -123 as x in g(x) to find our complete result

g{f[g(-4)]} = g(-123) = 8(-123) + 2 = -982

4 0

4 years ago

Which set of numbers cannot represent the lengths of the sides of a triangle?

igor_vitrenko [27]

I believe the answer would be d because 7+11 is not greater than 18

6 0

3 years ago

Refer to the following distribution of commissions:______. Monthly Commissions Class Frequencies $600 up to $800 3 $800 up to 1,

blsea [12.9K]

Answer:

the midpoint of the class with the greatest frequency 1500

Step-by-step explanation:

The illustration of the dataset given can be well represented in a table format as shown below.

Class Frequency Relative Frequency

$600 - $800 3 $\mathbf{ \dfrac{3}{120} = 0.025}$

$800 - $1000 7 $\mathbf{ \dfrac{7}{120} = 0.059}$

$1000 - $1200 11 $\mathbf{ \dfrac{11}{120} = 0.092}$

$1200 - $1400 22 $\mathbf{ \dfrac{22}{120} = 0.183}$

$1400 - $1600 40 $\mathbf{ \dfrac{40}{120} = 0.333}$

$1600 - $1800 24 $\mathbf{ \dfrac{24}{120} = 0.2}$

$1800 - $2000 9 $\mathbf{ \dfrac{9}{120} = 0.075}$

$2000 - $2200 4 $\mathbf{ \dfrac{4}{120} = 0.033}$

Total 120 1

Therefore, the midpoint of the class with the greatest frequency is between $1400 - $1600 since the frequency 40 happens to be the greatest frequency

The midpoint = $\dfrac{upper \ limit + \ lower \ limit}{2}$

The midpoint = $\dfrac{1400+1600}{2}$

= 1500

3 0

4 years ago

PLSSSS ANSWER I will give 35 POINTS

KengaRu [80]

Answer:

1.Radious and diameter of a circle

6 0

3 years ago

Read 2 more answers