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soldier1979 [14.2K]
3 years ago
6

If the remainders when three consecutive integers are each divided by 5 are 2, 3, and 4 respectively, which of the following cou

ld be the three integers? a.) 3, 4, 5 b.) 5, 6, 7 c.) 6, 7, 8 d.) 7, 8, 9 e.) 8, 9, 10 I don't understand please explain
Mathematics
1 answer:
mario62 [17]3 years ago
6 0
The answer is D. If you divide 7 by 5, you'll have a remainder of 2. If you divide 8 by 5, you'll have a remainder of 3, and if you divide 9 by 5 you'll have an remainder of 4. And respectively just means in order ..
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The first one is 810

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Step-by-step explanation: if I’m wrong just watch the video for a hint it helps

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Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ
Deffense [45]

Answer:

\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}

Step-by-step explanation:

Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}

\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

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\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i

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3 years ago
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Please helppp!!!! I’ll give you a crown!!
siniylev [52]

Answer:

the second one

Step-by-step explanation:

the last number of the equation (-3) is the y intercept, so it goes the the point (0,-3), also it has the opposite slope of y=2x-3

i hope this helps and im sorry if im wrong :(

8 0
3 years ago
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