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yuradex [85]
3 years ago
11

How many numbers are equal to the sum of two odd, one-digit numbers?

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

Seven numbers.

Step-by-step explanation:

Finding the numbers, which are equal to the sum of two odd number and it has to be single digit number.

Lets look into numbers which are odd and single digit.

1 = 1+3, 1+5, 1+7, 1+9

∴ Sum of the number is 4,6,8\ and\ 10

3 = 3+5, 3+7, 3+9

∴ Sum of above number is 8,10\ and\ 12

5 = 5+7, 5+9

∴ Sum of above number is 12\ and\ 14

7= 7+9

∴ Sum of above number is 16

Now, accumlating numbers which are fullfiling the criteria, however, making sure no number should get repeated.

∴ Numbers are: 4,6,8,10,12,14\ and\ 16

Hence, there are total 7 numbers, which  are equal to the sum of two odd, one-digit numbers.

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Find all the zeros of the equation x^4-6x^2-7x-6=0
rusak2 [61]

Answer:

The zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

a_0=6,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}

-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2

=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\

=x^3-2x^2-2x-3

Again factor using the rational root test, we get

=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)

Using the zero product rule, we have

x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Therefore, the zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}


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