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katrin [286]
3 years ago
7

Rearranging formulae:

Mathematics
2 answers:
bazaltina [42]3 years ago
7 0

Answer:

1. d/a+c=d

2. (m+21)/5=n

3. (1/2+2q)*4=p or 2+8q=p

4. (p-2a)/2pi=r

5. {[(5c+1)/2]+c}/3=a

stiv31 [10]3 years ago
5 0

Answer:

1. a (q - c ) = d

aq - ac = d

-ac - d = -aq

-ac - d = -aq

____ ___

-a -a

c - d÷a = q

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C or D as they are Identical

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Which type of triangle has one angle with a measurement greater than 90°?
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Answer:
Obtuse-angled triangle

Explanation:
Triangles can be classified based on the measures of their angles as follows:
1- Right-angled triangle: Triangle that has one of its angles = 90°
2- Acute-angled triangle: Triangle that has all of its angles less than 90°
3- Obtuse-angled triangle: Triangle that has one of its angles greater than 90°

The attached image shows an obtuse-angled triangle. We can note that measure ∠B is greater than 90°

Hope this helps :)

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How do you simplify cscx*secx-tanx?
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5 0
3 years ago
Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
3 years ago
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