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Sagot :
[tex]hw=315\\
h=w+6\\
(w+6)w=315\\
w^2+6w-315=0\\
w^2+21w-15w-315=0\\
w(w+21)-15(w+21)=0\\
(w-15)(w+21)=0\\
w=15 \vee w=-21\\\\
h=15+6=21\\\\\
w=15 \ in\\
h=21\ in\\[/tex]
Answer:
length of window = 21 in
width of window = 15 in
Explanation:
Assume that the width of the window is w
We are given that the length is 6 in longer than the width
This means that:
length of window = w + 6
Area of rectangle is calculated as follows:
area = width * length
area = w(w+6)
315 = w(w+6)
315 = w² + 6w
w² + 6w - 315 = 0
(w-15)(w+21) = 0
This means that:
either w-15=0 .......> w = 15 in .......> accepted solution
or w+21 = 0 .........> w = -21 in .......> rejected as width of window cannot be negative
Now, we know that:
length = width + 6
length = 15 + 6
length = 21 in
Based on the above calculations:
length of window = 21 in
width of window = 15 in
Hope this helps :)
length of window = 21 in
width of window = 15 in
Explanation:
Assume that the width of the window is w
We are given that the length is 6 in longer than the width
This means that:
length of window = w + 6
Area of rectangle is calculated as follows:
area = width * length
area = w(w+6)
315 = w(w+6)
315 = w² + 6w
w² + 6w - 315 = 0
(w-15)(w+21) = 0
This means that:
either w-15=0 .......> w = 15 in .......> accepted solution
or w+21 = 0 .........> w = -21 in .......> rejected as width of window cannot be negative
Now, we know that:
length = width + 6
length = 15 + 6
length = 21 in
Based on the above calculations:
length of window = 21 in
width of window = 15 in
Hope this helps :)
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