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lbvjy [14]
3 years ago
12

Please help I really need help on this I would really appreciate it anything helps !

Mathematics
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

<em>AXE</em> and <em>CXD  </em>are vertical angles

<em>AXF</em> and <em>FXD  </em>are supplementary angles

<em>DXC </em>and <em>BXC  </em>are complementary angles

<em>EXA</em> and <em>AXB  </em>are adjacent angles

<em>AXC </em>and <em>CXD</em>  are supplementary angles

<em>EXD </em>and <em>AXC </em>are vertical angles

Step-by-step explanation:

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Geometry, please answer question ASAP
Blababa [14]

9514 1404 393

Answer:

  D.  13.8

Step-by-step explanation:

FE being the midsegment means E is the midpoint of BC. The midpoint divides the segment into two equal parts, so BE = EC = 6.9. The length BC is the sum of those parts:

  BC = BE +EC = 6.9 +6.9 = 13.8

The length of BC is 13.8 units.

6 0
2 years ago
Whoever tells me what 10+10 is I’ll give out a brainliest to
bija089 [108]

Answer:

20

Step-by-step explanation:

7 0
3 years ago
What is the absolute deviation of 4 in this data set?
Nana76 [90]

Answer:

n

6

Mean

11

Average Absolute Deviation (MAD)

4.6667

Step-by-step explanation:

4 0
2 years ago
You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

6 0
3 years ago
Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64
tino4ka555 [31]

a.

x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}

By Fermat's little theorem, we have

x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5

x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7

5 and 7 are both prime, so \varphi(5)=4 and \varphi(7)=6. By Euler's theorem, we get

x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5

x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7

Now we can use the Chinese remainder theorem to solve for x. Start with

x=2\cdot7+5\cdot6

  • Taken mod 5, the second term vanishes and 14\equiv4\pmod5. Multiply by the inverse of 4 mod 5 (4), then by 2.

x=2\cdot7\cdot4\cdot2+5\cdot6

  • Taken mod 7, the first term vanishes and 30\equiv2\pmod7. Multiply by the inverse of 2 mod 7 (4), then by 6.

x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6

\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}

b.

x^5\equiv3\pmod{64}

We have \varphi(64)=32, so by Euler's theorem,

x^{32}\equiv1\pmod{64}

Now, raising both sides of the original congruence to the power of 6 gives

x^{30}\equiv3^6\equiv729\equiv25\pmod{64}

Then multiplying both sides by x^2 gives

x^{32}\equiv25x^2\equiv1\pmod{64}

so that x^2 is the inverse of 25 mod 64. To find this inverse, solve for y in 25y\equiv1\pmod{64}. Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that (-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}.

So we know

25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}

Squaring both sides of this gives

x^4\equiv1681\equiv17\pmod{64}

and multiplying both sides by x tells us

x^5\equiv17x\equiv3\pmod{64}

Use the Euclidean algorithm to solve for x.

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that (-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}, and so x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}

5 0
3 years ago
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