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Margaret [11]
3 years ago
15

What is m Please help me I’m willing to pay you for other problems just message me.

Mathematics
2 answers:
Blizzard [7]3 years ago
4 0

Answer:

56

Step-by-step explanation:

lesantik [10]3 years ago
4 0

56

hope this helps

¬_¬ :)

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I need to find the value of x with the given information
Vika [28.1K]

Answer:

BD is half of the answer BF is

Step-by-step explanation:

BD is only half of BF so find the answer for BF and then divide by @ and you have your answer.

6 0
3 years ago
∠1 and ∠2 are complementary angles. m∠1 is 5y+32 and m∠2 is 7y-14. Find m∠2 & show your work.
Ainat [17]
  • m<1+m<2=90

\\ \sf\longmapsto 5y+32+7y-14=90

\\ \sf\longmapsto 12y+18=90

\\ \sf\longmapsto 12y=72

\\ \sf\longmapsto y=6

  • m<2=7(6)-14=42-14=28[/tex]
7 0
3 years ago
Read 2 more answers
Mr. Musah withdrew some money from the bank. He gave ½ of it to his son and ⅓ to his daughter. if he had GH¢ 500.00 left, how mu
Savatey [412]

Answer:

GH 3000.00

Step-by-step explanation:

1/2 + 1/3 = 5/6

   6/6-5/6=1/6

  6/6 X 500.00 X6/1

 GH 3000.00

5 0
3 years ago
Can somebody help me please!!!!!
diamong [38]

Answer:

0

Step-by-step explanation:

g[f(x)] = x^2 - 3

= 3x^2

if gf(0) = 3(0)^2

= 0

3 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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