Answer:
55 days
Step-by-step explanation:
Given
Jim ran 15 miles in 5 days
no. of miles ran in 5 days = 15 miles
dividing LHS and RHS by 5
no. of miles ran in 5/5(=1) days = 15/5 miles = 3 miles
no. of miles ran in 1 day = 3 miles
let the no. of days taken to run 165 miles be x ----A
No of miles ran in x days = x*no. of miles ran in 1 day = 3x miles
thus, From A
3x = 165
x = 165/3 = 55
Thus, it took 55 days for JIM to run 165 miles
A) 4a - 3b + 2c
4(2, -1, 5) - 3(4, 3 , -2) + 2(5, 4, 0) = (8, -4, 20) - (12, 9, - 6) + (10, 8, 0) =
= (8 - 12 + 10 , -4 - 9 + 8 , 20 + 6 + 0) = (6, - 5, 26)
Answer: (6, - 5, 26)
b) magnitude of vector b
c) vector of length 7 parallel to vector c
=> m(5,4,0) = (5m,4m,0)
=>
=> m = 7 / √41 ≈ 1.093
=> 1.093 (5, 4, 0) = (5.465 , 4.372, 0)
Answer: (5.465 , 4.372 , 0)
Seven and Seventy-Five Hundredths
Answer: In the resulting equation: " a² - 12a + 32 = 0 " ;
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The "coefficient" of the "a" term is: " - 12" .
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The "constant" is: " 32 " .
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Explanation:
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Let: "a = x² + 4 " .
Given: (x² + 4)² + 32 = 12x² + 48 ;
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Factor: "12x² + 48" into " (x² + 4) " ;
"12x² + 48" = 12 (x² + 4) " ;
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Given: (x² + 4)² + 32 = 12x² + 48 ;
rewrite as; "a² + 32 = 12a " ;
Subtract "12a" from each side of the equation;
"a² + 32 - 12a = 12a - 12a ;
to get:
" a² - 12a + 32 = 0 " .
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The coefficient of the "a" term; that is:
The "coefficient" of " -12a" ; is: "- 12" .
The constant is: "32<span>" .
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