Answer
do u have a demonstration
picture
Step-by-step explanation:
Answer:
Step-by-step explanation:
![x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D](https://tex.z-dn.net/?f=x%5E2%2B3x%3D10%5E%7B1%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5Cx%5E2%2B3x%3D10%5E%7B1%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cqquad%5Ctext%7Buse%7D%5C%20a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C%5C%5Cx%5E2%2B3x%3D10%5Ccdot10%5E%5Cfrac%7B1%7D%7B2%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%5Cfrac%7B1%7D%7Bn%7D%5C%5C%5C%5Cx%5E2%2B3x%3D10%5Csqrt%7B10%7D%5Cqquad%5Ctext%7Bsubtract%7D%5C%2010%5Csqrt%7B10%7D%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5Cx%5E2%2B3x-10%5Csqrt%7B10%7D%3D0%5C%5C%5C%5C%5Ctext%7BUse%20the%20quadratic%20formula%7D%5C%5C%5C%5Cax%5E2%2Bbx%2Bc%3D0%5C%5C%5C%5Cx%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5Ca%3D1%2C%5C%20b%3D3%2C%5C%20c%3D-10%5Csqrt%7B10%7D%5C%5C%5C%5Cb%5E2-4ac%3D3%5E2-4%281%29%28-10%5Csqrt%7B10%7D%29%3D9%2B40%5Csqrt%7B10%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B-3%5Cpm%5Csqrt%7B40%2B10%5Csqrt%7B10%7D%7D%7D%7B2%281%29%7D%3D%5Cdfrac%7B-3%5Cpm%5Csqrt%7B40%2B10%5Csqrt%7B10%7D%7D%7D%7B2%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B-3-%5Csqrt%7B10%2B10%5Csqrt%7B10%7D%7D%7D%7B2%7D%5Cnotin%20D)
Answer:
hi there ☺️
Here we will use algebra to find three consecutive integers whose sum is 345. We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 345. Therefore, you can write the equation as follows:
(X) + (X + 1) + (X + 2) = 345
To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:
X + X + 1 + X + 2 = 345
3X + 3 = 345
3X + 3 - 3 = 345 - 3
3X = 342
3X/3 = 342/3
X = 114
Which means that the first number is 114, the second number is 114 + 1 and the third number is 114 + 2. Therefore, three consecutive integers that add up to 345 are 114, 115, and 116.
114 + 115 + 116 = 345
We know our answer is correct because 114 + 115 + 116 equals 345 as displayed above.
Step-by-step explanation:
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