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spin [16.1K]
3 years ago
6

Hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Mathematics
2 answers:
german3 years ago
7 0
9x9 = 81
Square root = 3
81-3 = 78
tatuchka [14]3 years ago
5 0
Hiiii? what is the question
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3(x + 1) + 12<br> 3<br>-I just need the anwser​
Goshia [24]
3x + 3 + 12
3x + 15
x + 5
3 0
3 years ago
Read 2 more answers
Given g(x)=x^2 and h(x)= x-3 find (g ∘ h)(x)
DerKrebs [107]

Answer:

<h2>(g ∘ h)(x) = x² - 6x + 9</h2>

Step-by-step explanation:

g(x) = x²

h(x) = x - 3

To find (g ∘ h)(x) substitute h(x) into g(x) that's for every x in g(x) replace it with h(x)

That's

(g ∘ h)(x) = ( x - 3)²

= x² - 3x - 3x + 9

We have the final answer as

<h3>(g ∘ h)(x) = x² - 6x + 9</h3>

Hope this helps you

5 0
3 years ago
Simplify 12p-3q +8q - 7p
IgorLugansk [536]

Answer:

5p+5q

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the interquartile range of 45,60,40,90,60,50,75,120,90
puteri [66]
IQR = 40

1) Put the numbers in order: 40, 45, 50, 60, 60, 75, 90, 90, 120
2) Find the median: Median is 60 (the 2nd one)
3) Place parentheses around the numbers above and below the median. For easy identification of Q1 and Q3. (40, 45, 50, 60,) 60, (75, 90, 90, 120)
4) Find the Q1 and Q3. Q1 = median of the lower half of the data; Q3 = median of the higher half of the data. Q1 and Q3 have even sets so its median cannot be defined.
5) Had both sets contain odd sets, the median of Q1 is subtracted from the median of Q3 to get the IQR.

We can then use the Alternative definition of IQR.
IQR is the difference between the largest and smallest values in the middle 50% of a set data.

40, 45, 50, 60, 60, 75, 90, 90, 120

Middle 50% is 50, 60, 60, 75, 90; IQR = Largest value - smallest value;

IQR = 90 - 50 = 40



5 0
3 years ago
Please help! will give brainliest !!
Veseljchak [2.6K]

Answer:

1, 2, 8, 128, 32768

Step-by-step explanation:

recursive \: formula: \\  a _{n} = 2. {( a_{n - 1} )}^{2}  \\  \\ a _{1} = 1...(given) \\  \\ a _{2} = 2. {( a_{2 - 1} )}^{2}   = 2 {(a _{1} )}^{2}  = 2 {(1)}^{2}  \\  \\  a _{2} = 2\\  \\a _{3} = 2. {( a_{3 - 1} )}^{2}   = 2 {(a _{2} )}^{2}  = 2 {(2)}^{2}  \\  \\  a _{3} = 8\\  \\a _{4} = 2. {( a_{4 - 1} )}^{2}   = 2 {(a _{3} )}^{2}  = 2 {(8)}^{2}  \\  \\  a _{4} = 128\\  \\a _{5} = 2. {( a_{5 - 1} )}^{2}   = 2 {(a _{4} )}^{2}  = 2 {(128)}^{2}  \\  \\  a _{5} = 32,768\\  \\

Thus the first five terms are: 1, 2, 8, 128, 32768.

5 0
3 years ago
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