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How do you find the width and length of a rectangular frame when the perimeter is 66in and the length is 3 in greater than the width?
Wiki User
∙ 11y ago
if the perimeter is 66in and the length is 3 all u have to do is this
3+3= 6
66-6=60
60/2=30
your width of one side of the rectangle is 30in
3
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30{ | | }30
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3
Wiki User
∙ 11y ago
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How do you get the length and width of a rectangular frame when the perimeter is 66 in The length is 3 in greater than the width?
The perimeter is the total of all the sides added up, which is two lengths and two widths. We can write the equation 66 = 2w + 2ℓ where 66 is perimeter, w is width, and ℓ is length. We can also write equation ℓ = 3w because the length is three times greater than the width. We can replace ℓ with 3w in the first equation because the second equation tells us they are the same thing: 66 = 2w + 2*(3w) Now we have an equation with one variable, so we can solve for that variable (w): 66 = 2w + 2*(3w) 66 = 2w + 6w 66 = 8w Divide both sides by 8: 8.25 = w The width is 6.6. We can plug that 6.6 into the equation ℓ = 3w to find the length: ℓ = 3*8.25 ℓ = 24.75 The length is 24.75. So the width is 8.25 and the length is 24.75.
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What happens to the mass of a speeding object in space?
Recall a fundamental postulate of relativity -- that one can not define the velocity of an object except in reference to a frame. Thus, we can NOT say an object is "speeding" unless we also define against which frame we are making measurements. In an object's own frame, its own mass never changes. In a frame that views such an object as "speeding," the mass of the object will be greater than it is in its own frame. Not "mistaken to be" greater, not "viewed as" greater, not "seems to be" greater, not "appears to be" greater. The mass IS greater in that second frame.
