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Yanka [14]
2 years ago
11

HEEEEEEEEEELP i dont get it

Mathematics
1 answer:
USPshnik [31]2 years ago
8 0

The ordered pair is (-9, -25) and the word statement is if x is equal to -9, then the value of h(x) is -25

<h3>Functions and tables</h3>

From the given table, f(x) = y means that the corresponding value of y given a value x.

For the function h(-9), we need to find the equivalent value of h(x) when x is -9. Hence h(-9) is -25

The ordered pair is (-9, -25) and the word statement is if x is equal to -9, then the value of h(x) is -25

Learn more on functions and values here: brainly.com/question/2284360

#SPJ1

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Angel has read 16 of the 80 pages in his book. What percent of the book has he read?
vovangra [49]

Answer:

Hey! so the answer is 20%

Step-by-step explanation:

If you want to know what percent 16 is of 80, you simple divide 16 by 80, then take that number and move the decimal place two spaces to the right. That's your percentage!

Hoped this helped! Have a great day! Im Eve btw. Also consider marking this brainliest. Thank you in advanced if you do!✨

6 0
3 years ago
Solve the following simultaneous linear congruences.
Anastaziya [24]

a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let

x=4\cdot5+3\cdot5+3\cdot4

  • Taken mod 3, the last two terms vanish, and 20\equiv2\pmod3 so we need to multiply by the inverse of 2 modulo 3 to end up with a remainder of 1. Since 2\cdot2\equiv4\equiv1\pmod3, we multiply the first term by 2.

x=4\cdot5\cdot2+3\cdot5+3\cdot4

  • Taken mod 4, the first and last terms vanish, and 15\equiv3\pmod4. Multiply by the inverse of 3 modulo 4 (which is 3 because 3\cdot3\equiv9\equiv1\pmod4), then by 2 to ensure the proper remainder is left.

x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4

  • Taken mod 5, the first two terms vanish, and 12\equiv2\pmod5. Multiply by the inverse of 2 modulo 5 (3, since 3\cdot2\equiv6\equiv1\pmod5) and again by 3.

x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4\cdot3\cdot3

\implies x=238

By the CRT, we have

x\equiv238\pmod{3\cdot4\cdot5}\implies x\equiv-2\pmod{60}\implies\boxed{x\equiv58\pmod{60}}

i.e. any number 58+60n (where n is an integer) satisifes the system.

b. The moduli are not coprime, so we need to check for possible contradictions. If x\equiv a\pmod m and x\equiv b\pmod n, then we need to have a\equiv b\pmod{\mathrm{gcd}(m,n)}. This basically amounts to checking that if x\equiv a\pmod m, then we should also have x\equiv a\pmod{\text{any divisor of }m}.

x\equiv4\pmod{10}\implies\begin{cases}x\equiv4\equiv0\pmod2\\x\equiv4\pmod5\end{cases}

x\equiv8\pmod{12}\implies\begin{cases}x\equiv0\pmod2\\x\equiv2\pmod3\end{cases}

x\equiv6\pmod{18}\implies\begin{cases}x\equiv0\pmod2\\x\equiv0\pmod3\end{cases}

The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.

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