+2446 I can simplify this for you!
| \((4b)^{-2}\) | First, let's use the rule that \(a^{-b}=\frac{1}{a^b}\). |
| \(\frac{1}{(4b)^2}\) | Simplify \((4b)^2\). To do this, "distribute the exponent to both the variable and the coefficient. |
| \(\frac{1}{4^2b^2}\) | Simplify 4^2. |
| \(\frac{1}{16b^2}\) | You cannot simplify any further. |
+2446 \((x-\sqrt{48})^2\leq\frac{49}{3}\)
I am sorry if I have interpreted your inequality incorrectly. I am fairly certain that the equation above is what you were attempting to convey.
| \((x-\sqrt{48})^2\leq\frac{49}{3}\) | Undo the square by square rooting both sides of the inequality. |
| \(x-\sqrt{48}\leq|\sqrt{\frac{49}{3}}|\) | Remember the square root rule that states that \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). I like to think of it as "distributing" the square root into both the numerator and denominator. Also, the square root function always results in a negative and positive answer. That's why the absolute value sign is there. |
| \(x-\sqrt{48}\leq|\frac{\sqrt{49}}{\sqrt{3}}|\) | Simplify the fraction on the right side of the equation. |
| \(x-\sqrt{48}\leq|\frac{7}{\sqrt{3}}|\) | Let's rationalize the right hand side of the equation by multiplying the numerator and denominator by |
| \(\frac{7}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\) | Reinsert this back into the inequality. |
| \(x-\sqrt{48}\leq|\frac{7\sqrt{3}}{3}|\) | The absolute value results in 2 answers: the positive and negative. Let's split that up. |
| \(x-\sqrt{48}\leq\pm\frac{7\sqrt{3}}{3}\) |
Now, we must evaluate each inequality separately. The one for the plus and the one for the minus. To make it easier, I'll solve both one at a time! I'll do the positive version first:
| \(x-\sqrt{48}\leq\frac{7\sqrt{3}}{3}\) | Add \(\sqrt{48}\) to both sides. |
| \(x\leq\frac{7\sqrt{3}}{3}+\sqrt{48}\) | First, I'll make\(\sqrt{48}\) into simplest radical form. |
| \(\sqrt{48}=\sqrt{16*3}=\sqrt{16}\sqrt{3}=4\sqrt{3}\) | Ok, because this radical is in simplest radical form, reinsert it back into the equation. |
| \(x\leq\frac{7\sqrt{3}}{3}+\frac{4\sqrt{3}}{1}\) | Put \(\frac{4\sqrt{3}}{1}\) into a common denominator so that it is possible to add them together. |
| \(\frac{4\sqrt{3}}{1}*\frac{3}{3}=\frac{12\sqrt{3}}{3}\) | Now that this fraction has a common denominator, let's add them together. |
| \(x\leq\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) | Add the fractions together. |
| \(x\leq\frac{19\sqrt{3}}{3}\) | Time to solve the second part of the equation. |
Now, we have to solve for the other scenario, \(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\). Now, this is a tad difficult to explain, but the absolute value function does funny things to an inequality. The rule is that \(|f(x)|\leq a\Rightarrow f(x)\leq a\hspace{1mm}\text{and}\hspace{1mm}f(x)\geq -a\).
| \(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\) | Solve and isolate x by adding \(\sqrt{48}\) to both sides. |
| \(x\geq-\frac{7\sqrt{3}}{3}+\sqrt{48}\) | Put \(\sqrt{48}\) into simplest radical form and put in common denomator. Because I have already done this in the previous equation, I'll just skip. |
| \(x\geq-\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) | Add the fractions together now. |
| \(x\geq\frac{5\sqrt{3}}{3}\) | |
Now, put the solutions together.
\(\frac{5\sqrt{3}}{3}\leq x\leq \frac{19\sqrt{3}}{3}\)
.+2446 Evaluating these algebraid expressions just requires you to substitute in the values given for the variable. Let's do the first one together:
1. \(N^2-25\)
a) When n=-10
| \(N^2-25\) | Replace N, the variable, with -10 and evaluate from there. |
| \((-10)^2-25\) | Do \((-10)^2=-10*-10=100\) first so to adhere to the order of operations. |
| \(100-25\) | |
| \(75\) | |
The process repeats for the other values for N.
b) n=-5
| \(N^2-25\) | |
| \((-5)^2-25\) | |
| \(25-25\) | |
| \(0\) | |
c) n=1/2
| \(N^2-25\) | |
| \((\frac{1}{2})^2-25\) | Remember that squaring a fraction follows the rule that \((\frac{a}{b})^2=\frac{a^2}{b^2}\) |
| \(\frac{1^2}{2^2}-25\) | Simplify the fraction. |
| \(\frac{1}{4}-25\) | Change 25 to an improper fraction so that you can subtract it from 1/4. |
| \(\frac{25}{1}*\frac{4}{4}=\frac{100}{4}\) | Now that we have changed 25 to a fraction with a common denominator, reinsert it back into the equation. |
| \(\frac{1}{4}-\frac{100}{4}\) | Subtract the fractions. |
| \(-\frac{99}{4}=-24\frac{3}{4}\) | The fraction is already in simplest form. I've provided both versions of the answer. |
d) n=9
| \(N^2-25\) | |
| \(9^2-25\) | |
| \(81-25\) | |
| \(56\) | |
I'll do half of the next one and the rest are up to you to complete:
2. \(\frac{-7d+14}{2}\)
a) d=2
| \(\frac{-7d+14}{2}\) | Substitute the given value for d, 2. |
| \(\frac{-7*2+14}{2}\) | Do -7*2 first. |
| \(\frac{-14+14}{2}\) | Simplify the numerator by calculating -14+14. |
| \(\frac{0}{2}=0\) | |
b) d=-2
| \(\frac{-7d+14}{2}\) | |
| \(\frac{-7*-2+14}{2}\) | A negative times a negative always results in a positive. |
| \(\frac{14+14}{2}\) | |
| \(\frac{28}{2}\) | Divide 28 by 2. |
| \(14\) | |
+2446 We can use long division to answer this question:
| 3x^2 | -7x | -4 | ||||||||
| ------ | ------ | ------ | ------ | ------ | ||||||
| | | 3x^3 | +2x^2 | -25x | -12 | ||||||
| x+3 | ------ | |||||||||
| - | (3x^3 | +9x^2) | ||||||||
| -3x^3 | -9x^2 | |||||||||
| ------ | --------- | |||||||||
| 0 | -7x^2 | -25x | ||||||||
| - | (-7x^2 | -21x) | ||||||||
| 7x^2 | +21x | |||||||||
| -------- | ------ | |||||||||
| 0 | -4x | -12 | ||||||||
| - | (-4x | -12) | ||||||||
| 4x | +12 | |||||||||
| 0 | 0 | |||||||||
Therefore, \(\frac{3x^3+2x^2-25x-12}{x+3}=3x^2-7x-4\hspace{1mm},\hspace{1mm}x\neq-3\)
.+2446 I'm unsure how this question is geometry-related. My only guess is that you are calculating the area of a triangle.
| \(\frac{1}{2}*5*3\) | When multiplying fractions together, I like to put integer values over 1. |
| \(\frac{1}{2}*\frac{5}{1}*\frac{3}{1}\) | Do \(\frac{1}{2}*\frac{5}{1}\) by multiplying the numerators and denominators. |
| \(\frac{5}{2}*\frac{3}{1}\) | Do \(\frac{5}{2}*\frac{3}{1}\) by multiplying the numerators and denominators. |
| \(\frac{15}{2}=7.5\) | |
There you go! I'm glad to help!
