We use the BODMAS method to solve the given questions.
Solve the Bracket then Divide Multiply Add and Subtract
1. 32÷4×(16×1/2) - 2
= 32÷4 × ( 8 ) - 2
= 8 × 8 - 2 = 64 - 2 = 62
2. 20 ÷ (5×2/5) + 3
= 20 ÷ (2) + 3 = 20 ÷ 2 + 3
= 10 + 3 = 13
3. 1/2 ×(6×4) ÷ 3 + 2
= 1/2 × 24 ÷ 3 + 2
= 1/2 × 8 + 2
= 4 + 2 = 6
4. 2 (6÷2+8) - 4
= 2 (3+8) - 4
= 2 × 11 - 4
= 22 - 4
= 18
the answer is 2 cups of peanut butter
Answer:
True
For the cube root function f(x)=3√x f ( x ) = x 3 , the domain and range include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function).
<em>Hope that helps!</em>
Step-by-step explanation:
First you should know : x ^ 0 = 1 and thats for all the numbers so m^0 = 1 too so now we just have : - n ^ 2 = - ( -1) ^2 = -1 so the answer is -1 :))
i hope this be helpful
* happy new year *
f(x)=x²−4x−21
The degree is the biggest power of x. That's a polynomial of degree 2, also called a quadratic function. Let's find its zeros.
0 = x²−4x−21 = (x - 7)(x+3)
x=7 or x=-3
The fundamental theorem guarantees every non-constant polynomial with complex coefficients has a complex zero, let's call it <em>r</em>. If we divide the polynomial by x-r there won't be any remainder and we'll get a new polynomial, one degree less. The fundamental theorem again applies and (if it's not a constant polynomial) we are assured of another zero, s. We divide by x-s and get a new polynomial of degree one less. We repeat all this until we get a constant polynomial (degree zero). So we get a zero for every degree. They're not necessarily all different.
Answer:
The degree of f(x) is 2. The Fundamental Theorem of Algebra guarantees that a polynomial equation has the same number of complex roots as its degree. This means that f(x) has exactly 2 zeros. Those zeros are 7 and -3.
