Let's solve your equation step-by-step.<span><span><span>4<span>x2</span></span>−48</span>=0</span>Step 1: Add 48 to both sides.<span><span><span><span>4<span>x2</span></span>−48</span>+48</span>=<span>0+48</span></span><span><span>4<span>x2</span></span>=48</span>Step 2: Divide both sides by 4.<span><span><span><span><span>4<span>x2</span></span>4</span></span></span>=<span><span><span>484</span></span></span></span><span><span>x2</span>=12</span>Step 3: Take square root.<span>x=<span>±<span>√12</span></span></span><span><span>x=<span><span><span><span><span>2<span>√3</span></span></span> or </span></span>x</span></span>=<span>−<span>2<span>√3</span></span></span></span>Answer:<span><span>x=<span><span><span><span><span>2<span>√3</span></span></span> or </span></span>x</span></span>=<span>−<span>2<span>√3</span></span></span></span>
I'm pretty new to expanded form, so this might be wrong, but I think it's:
(2 × 1/10) + (4 × 1/100)
You can also write it as:
0.2 + 0.04
Hope this helped!
Answer:
The ratio of bananas to apples is 9 : 4. The ratio of plums to apples is 3 to 4. For every 4 apples, there are 3 plums. For every 3 bananas there is one plum
You have to do the multiplication step, if there are any numbers being multiplied
Answer:
f'(x) > 0 on 
and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:
To find its decreasing interval :
2) Then let's find the critical point of this function:
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5B6-2%5E%7B2x%7D%5D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B6%5D-%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2%5E%7B2x%7D%5D%3D0-%5Bln%282%29%2A2%5E%7B2x%7D%2A%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2x%5D%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%3D-ln2%2A2%5E%7B2x%2B1%5CRightarrow%20%7Df%27%28x%29%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%5C%5C-ln%282%29%2A2%5E%7B2x%2B1%7D%3D-2x%5E%7B2x%7D%28ln%28x%29%2B1%29%3D0)
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''
≈0.37 for e≈2.72
3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
