it is a geometric situation and the formula is the following
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so after 6 hours we get
Answer:
(2x-1)(2x+1)(x^2+2) = 0
Step-by-step explanation:
Here's a trick: Use a temporary substitution for x^2. Let p = x^2. Then 4x^4+7x^2-2=0 becomes 4p^2 + 7p - 2 = 0.
Find p using the quadratic formula: a = 4, b = 7 and c = -2. Then the discriminant is b^2-4ac, or (7)^2-4(4)(-2), or 49+32, or 81.
Then the roots are:
-7 plus or minus √81
p= --------------------------------
8
p = 2/8 = 1/4 and p = -16/8 = -2.
Recalling that p = x^2, we let p = x^2 = 1/4, finding that x = plus or minus 1/2. We cannot do quite the same thing with the factor p= -2 because the roots would be complex.
If x = 1/2 is a root, then 2x - 1 is a factor. If x = -1/2 is a root, then 2x+1 is a factor.
Let's multiply these two factors, (2x-1) and (2x+1), together, obtaining 4x^2 - 1. Let's divide this 4x^2 - 1 into 4x^4+7x^2-2=0. We get x^2+2 as quotient.
Then, 4x^4+7x^2-2=0 in factored form, is (2x-1)(2x+1)(x^2+2) = 0.
To multiply decimals pretend there is no decimal for example
.165*74 = 12210; then however many digits (not counting zeros) is after the decimal added together (from both numbers) so 3 so 12.21 is the answer
20 because for all regular polygons their exterior angles add up to 360, so 360\18 =20
Answer:
Δ BEC ≅ Δ AED
Step-by-step explanation:
Consider triangles BCA and ADB. Each of them share a common side, AB. Respectively each we should be able to tell that AD is congruent to BC, and DB is congruent to CA, so by SSS the triangles should be congruent.
_________
So another possibility is triangles BEC, and AED. As you can see, by the Vertical Angles Theorem m∠BEC = m∠ADE, resulting in the congruency of an angle, rather a side. As mentioned before AD is congruent to BC, and perhaps another side is congruent to another in the same triangle. It should be then, by SSA that the triangles are congruent - but that is not an option. SSA does is one of the exceptions, a rule that is not permitted to make the triangles congruent. Therefore, it is highly unlikely that triangles BEC and AED are congruent, but that is what our solution, comparative to the rest.
Δ BEC ≅ Δ AED .... this is our solution