Answer:
3
Step-by-step explanation:
all numbers except from zero
that would mean 7 4 and 3
so there would be 3 significant figures
I hope it goes well
Answer:
would be A
Step-by-step explanation:
because bisecting means both sides equal the same. so 4x-12 times two is 8x-24. Hope this helps
The answer: m∡BCD = 130° .
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Explanation:
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m∡BCD = 9x - 5 = our answer.
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Note: (9x - 5) + (m∡C IN Δ ACB)= 180 ;
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Reason: all angles on straight line add up to 180.
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Note: In Δ ACB; m∡A + m∡B + m∡c = 180.
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Reason: All three angles in any triangle add up to 180.
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Given Δ ACB, we are given:
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m∡C= ?
m∡B = (4x + 5)
m∡A = 65
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So, given Δ ACB; m∡A + m∡B + m∡c = 180;
→Plug in our known values and rewrite:
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Given Δ ACB; 65 + 4x + 5 + (m∡c) = 180;
→Simplify, and rewrite:
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Given Δ ACB; 4x + 70 + (m∡c) = 180;
→Subtract "70" from each side of the equation; and rewrite:
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Given Δ ACB; 4x + (m∡C) = 110;
→Subtract "4x" from EACH SIDE of the equation; to isolate: "(m∡c)" on one side of the equation; and "solve in terms of "(m∡C)" ;
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Given Δ ACB' m∡C = 110 - 4x ;
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So, we know that: (110 - 4x) + (9x - 5) = 180; (since all angles on a straight line add up to 180.
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We can solve for "x".
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(110 - 4x) + (9x - 5) = 180;
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Rewrite as:
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(110 - 4x) + 1(9x - 5) = 180 ; (Note: there is an implied coefficient of "1"; since anything multiplied by "1" equals that same value).
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Note the "distributive property of multiplication":
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a*(b+c) = ab + ac ; AND:
a*(b - c) = ab - ac .
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So, +1(9x - 5) = (+1*9x) - (+1*5) = 9x - 5 ;
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So we can rewrite:
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(110 - 4x) + (9x - 5) = 180 ; as:
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110 - 4x + 9x - 5 = 180 ; We can simplify this by combining "like terms" on the "left-hand side" of the equation:
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110 - 5 = 105 ;
-4x + 9x = 5x;
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So, rewrite as: 5x + 105 = 180; Subtract "105" from EACH side; to get:
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5x = 75 ; Now, divide each side of the equation by "5";
to get: x = 15.
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Now, we want to know: m∡BCD; which equals:
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9x - 5 ; let us substitute "15" for "x"; and solve:
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9x - 5 = 9*(15) - 5 = 135 - 5 = 130.
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The answer: m∡BCD = 130°
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Answer:
i think the answer is C. $38.88
<u><em>hope it helps to your question!</em></u>
