What is the first integer to the right of 0 on the number line

Mathematics

1 answer:

Nesterboy [21]3 years ago

4 0

Answer:

Step-by-step explanation:

Numbers to the "right of 0" implies the positive numbers. And an integer has no fractional component. Thus, the first integer to the right of 0 would be 1.

Cheers.

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Which answer choice below correctly represents the quotient of 846 = 35?

avanturin [10]

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Step-by-step explanation:

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2 years ago

(x+16)²=12 plz help me and show work

Readme [11.4K]

Answer:

The answer is $x=-16$ ± $2\sqrt{3}$ in exact form or $x=-12.535898$ , $x=-19.464102$ in decimal form.

Step-by-step explanation:

To solve this problem, start by moving all terms to the left side of the equation and simplify. Simplify the equation by subtracting 12 from both sides of the equation and squaring $x+16$ , which will look like $x^{2} +32x+256-12=0$ . Next, simplify the equation again, which will look like $x^{2} +32x+244=0$ .

Then, use the quadratic formula to find the solutions. The quadratic formula looks like $\frac{-b(+-)\sqrt{b^{2}-4ac } }{2a}$ .

For this problem, the quadratic variables are as follows:

$a=1$

$b=32$

$c=244$

The next step is to substitute the values $a=1$ , $b=32$ , and $c=244$ into the quadratic formula and solve. The quadratic formula will look like $\frac{-32(+-)\sqrt{32^2-4(1*244)} }{2*1}$ . To simplify the equation, start by simplifying the numerator, which will look like $x=\frac{-32(+-)4\sqrt{3} }{2*1}$ . Then, multiply 2 by 1 and simplify the equation, which will look like $x=-16(+-)2\sqrt{3}$ . The final answer is $x=-16$ ± $2\sqrt{3}$ in exact form. In decimal form, the final answer is $x=-12.535898$ , $x=-19.464102$ .