Answer:
a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle
Step-by-step explanation:
Let l = the original length of the original rectangle
Let w = the original width of the original rectangle
From the description of the problem, we can construct the following two equations
l=2*w (Equation #1)
(l+4)*(w+4)=l*w+88 (Equation #2)
Substitute equation #1 into equation #2
(2w+4)*(w+4)=(2w*w)+88
2w^2+4w+8w+16=2w^2+88
collect like terms on the same side of the equation
2w^2+2w^2 +12w+16-88=0
4w^2+12w-72=0
Since 4 is afactor of each term, divide both sides of the equation by 4
w^2+3w-18=0
The quadratic equation can be factored into (w+6)*(w-3)=0
Therefore w=-6 or w=3
w=-6 can be rejected because the length of a rectangle can't be negative so
w=3 and from equation #1 l=2*w=2*3=6
I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.
Answer:
35°
Step-by-step explanation:
We assume that points S, T, U, V all lie on the circle.
Inscribed angle TUV intercepts long arc VST and short arc VUT. The long arc has measure 220°, so the measure of the short arc is 360° -220° = 140°.
Since chords UV and UT are the same length, point U bisects arc VUT. That means the measure of arc VU is 140°/2 = 70°.
Inscribed angle VSU intercepts arc VU, so has half the measure of the arc:
∠VSU = 70°/2
∠VSU = 35°
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Answer:
Step-by-step explanation:
