Let x be the initial width, then the initial length is x+20.
<span>If Elise decreases this length by 8 feet, she will get the new length x+20-8=x+12.
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If Elise increases the width by 10 feet, <span>she will get the new width x+10.
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</span><span>The new perimeter will be x+12+x+12+x+10+x+10=4x+44=172, 4x=172-44, 4x=128, x=128÷4, x=32.
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</span><span>The initial width is 32 ft and the initial length is 32+20=52 ft.
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</span><span>The new width will be 32+10=42 ft, the new length will be 32+12=44 ft.
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Answer: D
Step-by-step explanation:
Answer:
15 years old
Step-by-step explanation:
Start by defining the variables that we are going to use throughout our working:
Let the current age of Wei Ling and Wei Xuan be L and X years old respectively.
Next, form equations using the given information.
<u>5 years </u><u>ago</u>
Wei Ling: (L -5) years old
Wei Xuan: (X -5) years old
Given that the ratio of Wei Ling's age to that of Wei Xuan's is 2: 5,
Cross multiply:
2(X -5)= 5(L -5)
Expand:
2X -10= 5L -25
2X= 5L -25 +10
2X= 5L -15 -----(1)
<u>9 years time</u>
Wei Ling: (L +9) years old
Wei Xuan: (X +9) years old
Given that the ratio of Wei Ling's age to that of Wei Xuan is 3: 4,
Cross multiply:
3(X +9)= 4(L +9)
Expand:
3X +27= 4L +36
3X= 4L +36 -27
3X= 4L +9 -----(2)
Let's solve using the elimination method.
(1) ×3:
6X= 15L -45 -----(3)
(2) ×2:
6X= 8L +18 -----(4)
(3) -(4):
6X -6X= 15L -45 -(8L +18)
0= 15L -45 -8L -18
0= 7L -63
7L= 63
L= 63 ÷7
L= 9
Substitute L= 9 into (1):
2X= 5(9) -15
2X= 45 -15
2X= 30
X= 30 ÷2
X= 15
Thus, Wei Xuan is 15 years old now.
You should use Order of Operations or PEDMAS for this (Parenthesis, Exponents, Division and Multiplication in the order they appear, Addition and Subtraction in the order they appear). So we do the 2+12 because that is in parentheses first which gives 14-3+6. Next we do 14-3 because there are no exponents and no division or multiplication, which gives 11+6. This yields your answer, 17.
Answer:
<h2>|DB| = 6</h2>
Step-by-step explanation:
If ABCD is a rhombus then the diagonals AC and BD are perpendicular and intersect in half.
Let E be the point of intersection of the diagonals.
Therefore:
AE ≅ EC and DE ≅ EB ⇒ |AE| = |EC| and |DE| = |EB|.
DB = DE + EB
|EB| = 3 ⇒ |DB| = 3 + 3 = 6
ABCD it's not a rectangle because in each rectangle diagonals are congruent and intersect in half.