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Kay [80]
3 years ago
6

What number is a factor of 20 but not a multiple of 5

Mathematics
2 answers:
trasher [3.6K]3 years ago
7 0
A factor of 20 but not a multiple of 5 would probably be 4
Musya8 [376]3 years ago
7 0

Answer:

2 and 4.

Step-by-step explanation:

20 = 10*2

20 = 5*4

In the first example 2 is a factor of 20 but 2 is not a multiple of 5.

In the second example 4 is a factor of 20 but 4 is not a multiple of 5.

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5q greater than or equal to 8q -3/2
kakasveta [241]

5q ≥ 8q - 3/2

<em><u>Add 3/2 to both sides.</u></em>

3/2 + 5q ≥ 8q

<em><u>Subtract 5q from both sides.</u></em>

3/2 ≥ 3q

<em><u>Multiply both sides by 2.</u></em>

3 ≥ 6q

<em><u>Divide both sides by 6.</u></em>

0.5 ≥ q.

The value of q is less than or equal to 0.5.

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Use the remainder theorem to find the remainder for (x ^5 + 32) divided by (x+2) and state whether or not the binomial is a fact
marta [7]

x+2 is a factor of x^5+32 because (by the remainder theorem) the remainder upon dividing x^5+32 by x+2 is (-2)^5+32=0.

There's also the sum of fifth powers formula,

a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)

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3 years ago
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Image attachment below, Algebra 1 stuff
chubhunter [2.5K]
I know you didn't want an explanation, but I'll give you one anyway.
So, for a function, when you have f(5) = , then you'll put that 5 in for all of the x's in the function. 

The first batch of problems gave you the equation f(x) = \frac{x}{2} + 3. Now, plug in the numbers given for s and solve.

1. f(5) = <span>\frac{x}{2} + 3   Plug in 5 for x
</span>         = \frac{5}{2} + 3   Make three into a fraction over 2
         = \frac{5}{2} + <span>\frac{6}{2}   Add
         = </span><span><span>\frac{11}{2}

</span>2. f(-4) = </span><span>\frac{x}{2} + 3   Plug in -4 for x
           = </span><span>\frac{-4}{2} + 3   Divide -4 by 2
           = -2 + 3   Add
           = 1

For the next set, you have g(x) = x</span>² + 1.

3. Let's split this up. Solve the first equation, then the second, and then put         the answers together to solve it.

g(4) = x² + 1   Plug in 4 for x
       = 4² + 1   Square
       = 16 + 1   Add
       = 17

g(3) = x² + 1   Plug in 3 for x
       = 3² + 1   Square
       = 9 + 1   Add
       = 10

Now, put the two together.

g(4) + g(3) =    Substitute in the answers you just got.
      17 + 10 =    Add
                  = 27         

4. g(-1) = x² + 1      Substitute in -1 for x
            = (-1)² + 1   Square
            = 1 + 1        Add
            = 2

5. g(-6) = x² + 1       Plug in -6 for x
             = (-6)² + 1   Square
             = 36 + 1      Add
             = 37

For the last set, we have h(x) = x² + 7 and k(x) = 4x - 5. We'll have to pay close attention to the starting variables here.

6. h(-6) = x² + 7       Plug in -6 for x
             = (-6)² + 7   Square
             = 36 + 7      Add
             = 43

k(4) = 4x - 5     Plug in 4 for x
       = 4(4) - 5   Multiply
       = 16 - 5      Subtract
       = 11

Put the two answers into the final equation.

h(-6) + k(4) =    Substitute in the answers you got
       43 + 11 =    Add
                   = 54

7. k(10) = 4x - 5       Plug in 10 for x
            = 4(10) - 5   Multiply
            = 40 - 5      Subtract
            = 35

h(2) = x² + 7   Plug in 2 for x
       = 2² + 7   Square
       = 4 + 7     Add
       = 11

Now, put those into the equation.

k(10) - h(2) =    Plug in the answers you got
       35 - 11 =    Subtract
                  = 24

8. For this one, it wants you to add h(x) and k(x). We don't need to plug anything into the equations this time because they're both already solved for x! So, you can just set this up like a standard find-x equation.

(x² + 7) + (4x - 5) =
    x² + 7 + 4x - 5 =   Combine like terms (7 and -5)
         x² + 2 + 4x =    Reorder so the variables come first
                            = x² + 4x + 2

  
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