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Archy [21]
3 years ago
7

I don't get it someone please help me

Mathematics
2 answers:
Gre4nikov [31]3 years ago
7 0
1) 3 groups and 8 singles
2) All singles
3) Etc....
Alja [10]3 years ago
5 0
3 groups of 10 than 1 group of 8
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How do you work this out in confused?
Leokris [45]
I think you either need to double 35 or use a protractor. Hope this is helpful!
4 0
3 years ago
Read 2 more answers
Shureka Washburn has scores of 67​, 68​, 76​, and 63 on her algebra tests. a. Use an inequality to find the scores she must make
mash [69]
Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
(67+68+76+63+2x)/6≥71
(274+2x)/6≥71
solving the above we get:
274+2x≥71×6
274+2x≥426
2x≥426-274
2x≥152
x≥76

b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.


8 0
3 years ago
A flower pot's base has six sides that are all the same length. Each side measures x - 6 units. The base's perimeter is 78 units
Nadusha1986 [10]
The answer would x=19 units
5 0
3 years ago
One study found that the number of interest groups increased by what percentage between 1959 and 1995?
FrozenT [24]

Answer:

Percentage change in interest groups between 1959 (500) & 1995 (600) = 20%

Step-by-step explanation:

Percentage change = <u>Change ie (New - Old)  </u>  x 100

                                        Old value

Suppose -

Number of interest groups in 1959 = 500

Number of interest groups in 1995 = 600

Percentage change = [ (600 - 500) / 500 ] x 100

( 100 / 500 ) x 100 = 20%

6 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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