To get a fraction as a decimal, you have to divide the numerator by the denominator (commonly on a calculator, which is the easiest).
In this case, divide 20 by 100 to get the answer of 0.2.
So, 20/100 as a decimal is 0.2.
Whether dividing constant terms or polynomials, we always have definitive terms when it comes to division. Suppose we say, 10x divided by 2. The dividend is the 10x and the divisor is the 2. In other words, the dividend is the number to be divided by the divisor, to obtain the answer called the quotient.
When dividing polynomials, your main goal is to be able to divide the dividend evenly into the <em>divisor</em>. For example, we divide x²+2x+1 by x+1. The first thing you're going to focus is, what term will completely divide the first term of the polynomial? That would be x. Why? Because when you multiply x with x+1, the product is x²+x. When you subtract this from the polynomial, the x² will cancel out. All you have to do is subtract x from 2x, yielding x. Then, you carry down the last term of the equation: +1. You do the steps again. The term that will completely divide x+1 by x+1 is 1. When you subtract the two, you will come up with zero. That means there is no remainder. The polynomial is divisible by the divisor.
x + 1
------------------------------------
x+1| x²+2x+1
- x²+x
----------------------
x +1
- x +
------------
0
If two figures are similar, the resulting corresponding angles are always congruent.
Answer:
-1 and -4
Step-by-step explanation:
-1 + -4 = -5
-1 times -4 = 4
give brainiest please!
hope this helps ;)
Answer:
Final answer is ![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
.
Step-by-step explanation:
Given problem is ![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
.
Now we need to simplify this problem.
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
Hence final answer is .
