Answer:
x=2
Step-by-step explanation:
Answer:
$108
Step-by-step explanation:
<u>Use the equation i = prt </u><u>(interest = principle*rate*time)</u><u>:</u>
Principle: $900
Rate: 0.06 (6%)
Time: 2 years
<u>Put it all in an equation:</u>
i = 900*0.06*2
i = $108
V=-4 this should be the answer
Answer:
<ACB is supplementary to the < ECD by the definition of supplementary angles. Therefore the first part of your answer is <ACB = <ECD
We know that similar triangles have the same corresponding angle measurements for their angles and therefore one of the triangles is just a scaled up or down version of the original. Triangle EDC is a scaled down version of ABC. Therefore, the corresponding sides of triangle ECD have the same ratio to the corresponding sides of ABC. For example line segment EC and BC are corresponding line segment and lets find the ratio: 14/21 = 2/3
THe ratio of EC to BC is 2/3 and the ratio of CD to AC is also 2/3 (18 / 27=2/3)
Therefore, CE / BC is equal to DC / AC
Your full answer is
<ACB = <ECD, CE / BC = DC / AC
So triangle ABC ~ triangle CDE
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)
