Given: f(x) = 3x-7, g(x)=2x2 -3x+1, h(x)=4x+1, k(x) =-x2+3 Find: k(x) +g(x)

Mathematics

1 answer:

Afina-wow [57]3 years ago

7 0

Answer:

k(x) + g(x) = x² - 3x + 4

General Formulas and Concepts:

Alg I

Combining like terms

Step-by-step explanation:

Step 1: Define

f(x) = 3x - 7

g(x) = 2x² - 3x + 1

h(x) = 4x + 1

k(x) = -x² + 3

Step 2: Find k(x) + g(x)

Substitute: k(x) + g(x) = -x² + 3 + 2x² - 3x + 1
Combine like terms (x²): k(x) + g(x) = x² + 3 - 3x + 1
Combine like terms (Z): k(x) + g(x) = x² - 3x + 4

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Last year a baseball team paid $20 per bat and $12 per glove spending a total of $1040 this year the prices went up to $25 per b

Sonja [21]

20x + 12y = 1040
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-12y' to each side of the equation. 20x + 12y + -12y = 1040 + -12y Combine like terms: 12y + -12y = 0 20x + 0 = 1040 + -12y 20x = 1040 + -12y Divide each side by '20'. x = 52 + -0.6y
thats the first part
then we have
25x + 16y = 1350
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-16y' to each side of the equation.
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7 0

3 years ago

Find the following Euler Totients using Euler’s Theorem, as explained on p.409 of the text (10 points each): a.ϕ(13) b.ϕ(81) c.ϕ

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(a) $\varphi(13)=12$ since 13 is prime.

(b) $81=3^4$ , and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so $\varphi(81)=54$ .

(c) $100=2^2\cdot5^2$ ; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with $\varphi(100)=100-60=40$ .

(d) $102=2\cdot3\cdot17$ ; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so $\varphi(102)=102-70=32$ .