What’s the square root of 12?
2 answers:
Answer:
2√3
Step-by-step explanation:
√2^2x3
√2^2 √3
2√3 (answer)
Answer:
![\large\boxed{\sqrt{12}=2\sqrt3\approx3.46}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Csqrt%7B12%7D%3D2%5Csqrt3%5Capprox3.46%7D)
Step-by-step explanation:
![\sqrt{12}=\sqrt{4\cdot3}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt4\cdot\sqrt3=2\sqrt3\\\\\text{If you want to get an approximate value, use the calculator:}\\\\\sqrt{12}\approx3.46](https://tex.z-dn.net/?f=%5Csqrt%7B12%7D%3D%5Csqrt%7B4%5Ccdot3%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%5Csqrt%7Bab%7D%3D%5Csqrt%7Ba%7D%5Ccdot%5Csqrt%7Bb%7D%5C%5C%5C%5C%3D%5Csqrt4%5Ccdot%5Csqrt3%3D2%5Csqrt3%5C%5C%5C%5C%5Ctext%7BIf%20you%20want%20to%20get%20an%20approximate%20value%2C%20use%20the%20calculator%3A%7D%5C%5C%5C%5C%5Csqrt%7B12%7D%5Capprox3.46)
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Dividing 20x - 16xy by 4x gives 5 - 4y
so the product is
4x(5 - 4y)
Answer:
![\large\boxed{B.\ x=1,\ C.\ x=-\dfrac{5}{3}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7BB.%5C%20x%3D1%2C%5C%20C.%5C%20x%3D-%5Cdfrac%7B5%7D%7B3%7D%7D)
Step-by-step explanation:
![2x-5=-3x^2\qquad\text{add}\ 3x^2\ \text{to both sides}\\\\3x^2+2x-5=0\\\\3x^2+5x-3x-5=0\\\\x(3x+5)-1(3x+5)=0\\\\(3x+5)(x-1)=0\iff3x+5=0\ \vee\ x-1=0\\\\3x+5=0\qquad\text{subtract 5 from both sides}\\3x=-5\qquad\text{divide both sides by 3}\\x=-\dfrac{5}{3}\\\\x-1=0\qquad\text{add 1 to both sides}\\x=1](https://tex.z-dn.net/?f=2x-5%3D-3x%5E2%5Cqquad%5Ctext%7Badd%7D%5C%203x%5E2%5C%20%5Ctext%7Bto%20both%20sides%7D%5C%5C%5C%5C3x%5E2%2B2x-5%3D0%5C%5C%5C%5C3x%5E2%2B5x-3x-5%3D0%5C%5C%5C%5Cx%283x%2B5%29-1%283x%2B5%29%3D0%5C%5C%5C%5C%283x%2B5%29%28x-1%29%3D0%5Ciff3x%2B5%3D0%5C%20%5Cvee%5C%20x-1%3D0%5C%5C%5C%5C3x%2B5%3D0%5Cqquad%5Ctext%7Bsubtract%205%20from%20both%20sides%7D%5C%5C3x%3D-5%5Cqquad%5Ctext%7Bdivide%20both%20sides%20by%203%7D%5C%5Cx%3D-%5Cdfrac%7B5%7D%7B3%7D%5C%5C%5C%5Cx-1%3D0%5Cqquad%5Ctext%7Badd%201%20to%20both%20sides%7D%5C%5Cx%3D1)
11b. Is the answer. Okokok
Answer:
56 because you have to multiply 28 times 2
Answer:
55555555
Step-by-step explanation: