The option that can be used to support the idea that the set of polynomials is closed under multiplication is; Option C: (10x^(0.5) - 8)(5x^(0.5) + 4)
<h3>What is the Closure property under multiplication?</h3>
When multiplying polynomials, the variables' exponents are added, according to the rules of exponents. It is pertinent to note that the exponents in polynomials are whole numbers. The whole numbers are closed under addition, which guarantees that the new exponents will be whole numbers. Thus, we can also say that the polynomials are closed under multiplication.
Now, looking at the options, we can say that option C is the only polynomial that is closed under multiplication because its' variables and exponents will not change;
(10x^(0.5) - 8)(5x^(0.5) + 4)
The output will retain the same thing.
Read more about closure property at; brainly.com/question/19340450
#SPJ1
Answer:
Random sample, 
and 
, so yes, both conditions were satisfied.
Step-by-step explanation:
60% of the people in a certain midwest city who are responsible for preparing the evening meal have no idea what they are going to prepare as late as 4PM in the afternoon.
This means that 
A recent survey was conducted from 1000 of these individuals.
This means that 
Also, a random sample, so the first condition was satisfied.
The sample size must be large (so that at least 10 or more successes and failures).
So yes, both conditions were met.
Answer:
x=−27/26
and y=−192/13
Step-by-step explanation:
6x+y=−21;8x−3y=36
Step: Solve6x+y=−21for y:
6x+y=−21
6x+y+−6x=−21+−6x(Add -6x to both sides)
y=−6x−21
Step: Substitute−6x−21foryin8x−3y=36:
8x−3y=36
8x−3(−6x−21)=36
26x+63=36(Simplify both sides of the equation)
26x+63+−63=36+−63(Add -63 to both sides)
26x=−27
26x
26
=
−27
26
(Divide both sides by 26)
Answer:
-1.92 cm³
Step-by-step explanation:
The formula for the volume of a cube in terms of side length is ...
V = s³
Then the first-order approximation of volume for a small change ∆s in side length is ...
V(s +∆s) ≈ V(s) + V'(s)·∆s
and the change in volume is ...
∆V ≈ V(s +∆s) -V(s) = V'(s)·∆s
The derivative V'(s) is ...
V'(s) = 3s²
so the change in volume for s = 8 cm and ∆s = -0.01 cm is ...
∆V ≈ 3s²·∆s = 3(8 cm)²(-0.01 cm) = -1.92 cm³
It
can be described useing two other undefined terms :- point & line
