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kramer
3 years ago
5

Please answer #4 A-C!

Mathematics
1 answer:
UNO [17]3 years ago
3 0

Answer:

Hey, I want to help but I am not able to see the image. Please message me.

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Please helpp me (and explain if you can pls:) ) TYSMMM!!!!
Vedmedyk [2.9K]

Answer:

the last three

Step-by-step explanation:

cuz im big brains like that

5 0
2 years ago
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15% of 25 is what? Please answer asap
olganol [36]

Answer: 3.75

Step-by-step explanation:

5 0
2 years ago
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given the functions f(x)=4x^2-1, g(x)=x^2-8x+5, and h(x)=-3x^2-12x +1, rank them from least to greatest based on their axis of s
Lesechka [4]
The formula for axis of symmetry is -b/2a.

For f(x):
0/2(4)
0/8
Axis of symmetry= 0

For g(x)
-(-8)/2(1)
8/2
Axis of symmetry= 4

For h(x)
-(-12)/2(-3)
12/-6
Axis of symmetry= -2

Order from least to greatest: -2, 0, 4

Final answer: h(x), f(x), g(x)
7 0
2 years ago
Frank made a New Years resolution to get into better shape. He decides to join LA fitness. He has to pay a one-time enrollment f
professor190 [17]

Answer: an equation that represents the total costs of the gym membership based on the number of months is

y = 25x + 50

Step-by-step explanation:

Let x represent the number of months that Frank makes use of the gym at LA fitness in order to get better in shape.

Let y represent the total cost of using the gym for x months.

He has to pay a one-time enrollment fee of $50 and then membership costs $25 per month. This means that the total cist for x month would be

y = 25x + 50

8 0
3 years ago
*20 POINTS*
Paraphin [41]

Answer:

If you want to use the Rational Zeros Theorem, as instructed, you need to use synthetic division to find zeros until you get a quadratic remainder.

P: ±1, ±2, ±3, ±6  (all prime factors of constant term)

Q: ±1, ±7              (all prime factors of the leading coefficient)

P/Q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7  (all possible values of P/Q)

Now, start testing your values of P/Q in your polynomial:

f(x)=7x4-9x3-41x2+13x+6

You can tell f(1) and f(-1) are not zeros since they're not = 0Now try f(2) and f(-2):

f(2)=7(16)-9(8)-41(4)+13(2)+6

       112-72-164+26+6 ≠ 0

f(-2)=7(16)-9(-8)-41(4)+13(-2)+6

       112+72-164-26+6 = 0  OK!! There is a zero at x=-2

This means (x+2) is a factor of the polynomial.

Now, do synthetic division to find the polynomial that results from

(7x4-9x3-41x2+13x+6)÷(x+2):

-2⊥ 7   -9   -41    13     6

         -14   46   -10    -6          

      7  -23   5       3     0     The remainder is 0, as expected

The quotient is a polynomial of degree 3:

7x3-23x2+5x+3

Now, continue testing the P/Q values with this new polynomial.  Try f(3):

f(3)=7(27)-23(9)+5(3)+3

      189-207+15+3 = 0  OK!!  we found another zero at x=3

Now, another synthetic division:

3⊥ 7   -23    5     3

          21   -6   -3_

    7    -2    -1    0  

The quotient is a quadratic polynomial:

7x2-2x-1  This is not factorable, you need to apply the quadratic formula to find the 3rd and 4th zeros:

x= (1±2√2)÷7

The polynomial has 4 zeros at x=-2, 3, (1±2√2)÷7

Step-by-step explanation:

7 0
3 years ago
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