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laila [671]
3 years ago
13

HELP ASAP PLSSS

Mathematics
1 answer:
pogonyaev3 years ago
6 0

Answer:

complementry and ajaecent

Step-by-step explanation:

complementry because they add up to 90 and ajecent cause they are next to each other

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Which formula would be used to find the measure of angle 1?
ankoles [38]

Answer:

I believe its the 4th option on edge...

Step-by-step explanation:


7 0
3 years ago
Read 2 more answers
A company issues 7% bonds with a par value of $200,000 at par on January 1. The market rate on the date of issuance was 6%. The
Step2247 [10]

Answer:

The cash  paid to the bondholder on July 1 is    Z =  $7000

Step-by-step explanation:

From the question we are told that

  The percentage bond issued by the company is n = 7%

    The par value of the bond is V =$200,000

     The market rate is r = 6%

So we are told that the bonds pay interest semiannually on January 1 and July

So the cash  paid to the bondholder on July 1 is mathematically evaluated as

Z =  V * \frac{7}{100} * \frac{1}{2}        

substituting value

   Z =  200000 * \frac{7}{100} * \frac{1}{2}  

      Z =  $7000

3 0
4 years ago
Given the sequence 8, 16, 32, 64, ..., which expression would give the thirteenth term?
mrs_skeptik [129]
8, 16, 32, 64,…
 ×2  ×2  ×2

a_{n} = a_{1} * r^{n - 1}
a_{n} = 8 * 2^{13 - 1}
a_{n} = 8 * 2^{12}
a_{n} = 8 * 4096
a_{n} = 32768

           First Term: 8
      Second Term: 16
          Third Term: 32
        Fourth Term: 64
           Fifth Term: 128
          Sixth Term: 256
     Seventh Term: 512
       Eighth Term: 1024
         Ninth Term: 2048
        Tenth Term: 4096
   Eleventh Term: 8192
     Twelfth Term: 16384
Thirteenth Term: 32768

The\ expression\ that\ would\ give\ the\ thirteenth\ term\ of\ the\ problem\ would\ be\ the\ equation\ of\ the\ geometric\ sequence,\ which\ is\ equal\ to\ a_{n} = a_{1} * r^{n - 1}.

5 0
3 years ago
Read 2 more answers
50 points.
Sphinxa [80]

Answer:

1

Step-by-step explanation:

Using the rules of exponents

(a^m)^{n} = a^{mn}

a^{m} × a^{n} = a^{(m+n)}

Given

(\frac{1}{x^{(a-b)}) } ^{a+b} × (\frac{1}{x^{b+a}) } ^{b-a}

= \frac{1}{x^{(a-b)(a+b)} } × \frac{1}{x^{(b+a)(b-a)} }

= \frac{1}{x^{(a^2-b^2)} } × \frac{1}{x^{(b^2-a^2)} }

= \frac{1}{x^{(a^2-b^2+b^2-a^2)} }

= \frac{1}{x^{0} }

= \frac{1}{1}

= 1

5 0
3 years ago
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Condense log2 2 + 4 log2 as a single logarithm.
zysi [14]
Just condone it and multiply it with the algorithm.
3 0
3 years ago
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