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With fractional exponents you can go about it either way as long as you keep the same principal with regards to how you handle it!
The numerating exponent is what you exponentiate, while the denominator is what you root.
So with this case,
27^2/3 = 3√(27^2)
So we could do that first method but it gives us a huge number to cube root: 3√729 = 9
But let’s assume we’re doing this without a calculator, so the second method becomes easier.
It’s first asking for the cube root of 27 which if you don’t know what rooting means, basically asks for what one single number raised to the power of 3 would give us 27:
_ * _ * _ = 27 or ___^3 = 27
Now if you know your multiples of three you could see that 3 x 3 x 3 = 27, so that the cube root of 27 is 3.
So now you have
Which now is simply 3 squared (3x3 = 9)
Hope this helps!
I’m going to assume you’re asking how to find the roots of this equation.
First, let’s make the equation 2(x-2)^2 = x-2 a little less messy. You see that (x-2) is on both sides. Thus, you can divide both sides by (x-2), obtaining:
2(x-2) = 1
We can make the left side of the equation more clear by distributing that 2 into the parentheses, getting:
2x-4 = 1
Now, to get roots, you want all the variables on one side of the equals sign, and everything else on the other side. So let’s add 4 to both sides to get:
2x = 5
Now, we only want what x equals, so we divide both sides by 2.
x = 5/2.
I’m going to include a few examples and a (hopefully) detailed description, so here we go:
(I’m going to assume this is using calculus, because I can’t think of a fast way than that right now… Send us another message if it isn’t!)
So this is the equation we’re given:
y = x^4 - 1
Now we need to find the slope, by differentiation:
dy/dx = 4x^3
We know that the slope is equal to 32 at point p, which is what we want:
32 = 4x^3
Then we can just solve for x!
8 = x^3
x = 2
I hope this helps :)
When you’re looking for zeros of polynomials and you already have one of the results, you should try Horner’s method(/scheme/rule).
It works like this: write the coefficients of your polynomial in one row (ALL of them. For example, if one of the coefficients is 0, you will need to write that down, too.)
In your case, it would be 1 -3 1 1.
Now you would write the number you’re suspecting to be a zero in the next row, but to the left so it looks like a kind of table now.
Then you start calculating. You always add the numbers downwards, and to get to the next column you mutiply by your zero suspect. So it’d be
1 -3 1 1 -> top row: coefficients of your origional polynomial
1 / 1 -2 -1 -> middle row: zero suspect and calculation
1 -2 -1 0 -> bottom row: coefficients of your resulting poynomial
In case your zero suspect turn out to actually be a zero of your polynomial, the last number on the bottom right should be a 0. In our case, it is, so the bottom row now tell us which coefficients our resulting polynomial has.
We have now divided x³-3x²+x+1 by x-1 (the first zero) and got x²-2x-1.
You are right, the other two zeros aren’t whole numbers but you can use the regular methods for finding roots of quadratic equations. (you could, for example, check our tags …)
I hope that helped!
28 - (-(-15)) - |10| = ?
The double negatives cancel inside the bracket and the | | just means that the number inside becomes positive, since it’s already positive the absolute sign drops to make:
28 - (15) - 10 = ?
13 - 10 = 3
To find each intercept, you must let the other variable be equal to zero to find the point of the other axis.
For example, if you were to find the x intercept on this function you allow:
y = 0
3x + 4(0) = -12
3x + 0= -12
x = -4
and to find the y intercept you let x =0
3(0) + 4y = -12
0 + 4y =-12
y = -3
so you’re points of intercept are (-4,0) for the x intercept. and (0,-3) for the y intercept
Hi! Sorry for the delay in answering your question! Here’s a solution to the problem, I even did a little extra to show you how you can develop an idea of how the function would look when plotted.
If you need a higher quality picture try this one here