Isiah is 5.83 miles away from his starting point.
When you draw the the miles Isaiah walks, you will discover that you are forming a right angle triangle.
To measure the length from the starting point to the ending point. We must used the Pythagorean equation. Where in the square of the hypotenuse is equal to the sum of squares of the other two sides.
a² + b² = c²
a = 3 miles 3² + 5² = c²
b = 5 miles 9 + 25 = c²
c = hypotenuse. 34 = c²
We must get the hypotenuse, since 34 is the square of the hypotenuse we must look for its square root.
c² = 34
c = √34
c = 5.83 miles
1 yard = 3 feet.
Divide total feet by 3 to get yards:
72 feet / 3 = 24 yards.
Hello from MrBillDoesMath!
Answer:
The first choice ( -12, 20)
Discussion:
Using matrix multiplication we find the equations:
1x + 1y = 8 (A)
2x + 3y = 36 (B)
Next, compute
2(A) - B =>
(2x + 2y) - (2x + 3y) = 2*8 - 36 =>
(2x - 2x) + ( 2y-3y) = 16 - 36 =>
-y = -20 =>
y = 20
Substituting this y value in (A) gives
x + 20 = 8 =>
x = 8 - 20 =>
x = -12
The answer is the first choice ( -12, 20)
Thank you,
MrB
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Given:
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y = - 4x + 16 ;
4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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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 ;
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So, we can now rewrite the two (2) equations in the given system:
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y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;
4y − x = -4 ; ===> Refer to this as "Equation 2" ;
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Solve for "x" and "y" ; using "substitution" :
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We are given, as "Equation 1" ;
→ " y = - 4x + 16 " ;
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→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;
to solve for "x" ; as follows:
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Note: "Equation 2" :
→ " 4y − x = - 4 " ;
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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:
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→ " 4 (-4x + 16) − x = -4 " ;
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Note the "distributive property" of multiplication :
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a(b + c) = ab + ac ; AND:
a(b − c) = ab <span>− ac .
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As such:
We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
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AND:
→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
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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 ;
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Now, Plug this value for "x" ; into "{Equation 1"} ;
which is: " y = -4x + 16" ; to solve for "y".
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→ y = -4(4) + 16 ;
= -16 + 16 ;
→ y = 0 .
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The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Now, let us check our answers—as directed in this very question itself ;
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→ 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!
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→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
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→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
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Hope this lenghty explanation is of help! Best wishes!
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