Answer:
Therefore,
P = { 1 , 2 , 3 , 4 , 5 }
Step-by-step explanation:
Natural numbers:
Natural numbers are those numbers starting from 1 , 2 , 3 , 4 ,......... and so on.
Also it is denoted by 'N'.
Zero does not include a natural number.
Therefore the set of natural numbers less than 6 is 1 , 2 , 3 , 4 , and 5.
Here in the question it is given that
P is the set of natural numbers
less than 6.
∴ P = { 1 , 2 , 3 , 4 , 5 }
The prime factorisation for 58 is 2 x 29
:)
Answer:
They would need approximately 20 gallons of paint for the water tower, so yes, 25 gallons would be enough.
Step-by-step explanation:
In order to find out if there is enough paint for the surface area of the sphere, you need to first find the radius of the object so you can determine the measure of the entire surface area. Since they give us the volume, we can use that to solve for the radius using the volume formula: V = πr³ or 66840.28 = πr³. Using inverse operations, you can first multiply both sides of the equation by the reciprocal of 4/3 or 3/4, then divide by π, this gives you 15,965 = r³. In order to find 'r', take the cube root of both sides: ∛15965 = ∛r³ or r ≈ 25.
Now that you know radius = 25, you can use the formula for surface area: SA = 4πr² or 4π(25)² ≈ 7850 ft². Since one gallon of paint covers 400 square feet, we can take our total and divide by 400 to find the number of gallons: 7850 ÷ 400 ≈ 19.6. In total, we would need 19.6 gallons of paint for the water tower, so if there are 25 gallons available, there is enough.
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;
4y − x + 4 = 0 ;
________________________________________________________
"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ;
→ 4y − x + 4 = 0 ;
Subtract "4" from each side of the equation ;
→ 4y − x + 4 − 4 = 0 − 4 ;
→ 4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________
y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;
4y − x = -4 ; ===> Refer to this as "Equation 2" ;
________________________________________________________
Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;
→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;
to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :
→ " 4y − x = - 4 " ;
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;
→ as follows:
_________________________________________________
→ " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________
a(b + c) = ab + ac ; AND:
a(b − c) = ab <span>− ac .
_________________________________________________
As such:
We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:
→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:
→ " -16x + 64 − x = - 4 " ;
Note: " - 16x − x = -16x − 1x = -17x " ;
→ " -17x + 64 = - 4 " ; Solve for "x" ;
Subtract "64" from EACH SIDE of the equation:
→ " -17x + 64 − 64 = - 4 − 64 " ;
to get:
→ " -17x = -68 " ;
Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -17x / -17 = -68/ -17 ;
to get:
→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;
which is: " y = -4x + 16" ; to solve for "y".
______________________________________
→ y = -4(4) + 16 ;
= -16 + 16 ;
→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;
→ Let us check;
→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;
→ Consider the first equation given in our problem, as originally written in the system of equations:
→ " y = - 4x + 16 " ;
→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?
→ "0 = ? -4(4) + 16 " ?? ; → "0 = ? -16 + 16 ?? " ; → Yes! ;
{Actually, that is how we obtained our value for "y" initially.}.
→ Now, let us check the other equation given—as originally written in this very question:
→ " 4y − x + 4 = ?? 0 ??? " ;
→ Let us "plug in" our obtained values into the equation;
{that is: "4" for the "x-value" ; & "0" for the "y-value" ;
→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.
→ " 4(0) − 4 + 4 = ? 0 ?? " ;
→ " 0 − 4 + 4 = ? 0 ?? " ;
→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________
Oh okie so I Pepeooooo oit the way
