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.
Answer:
it is 12
Step-by-step explanation:
Answer:
-6
Step-by-step explanation:
You need to get it in Point-Slope form, or
y = mx + b
Isolate y by subtracting 6x
You are left with:
y = -6x + 24
M is the slope, so in this case, it is -6
Answer: The total number of logs in the pile is 6.
Step-by-step explanation: Given that a stack of logs has 32 logs on the bottom layer. Each subsequent layer has 6 fewer logs than the previous layer and the top layer has two logs.
We are to find the total number of logs in the pile.
Let n represents the total number of logs in the pile.
Since each subsequent layer has 6 fewer logs then the previous layer, so the number of logs in each layer will become an ARITHMETIC sequence with
first term, a = 32 and common difference, d = -6.
We know that
the n-th term of an arithmetic sequence with first term a and common difference d is

Since there are n logs in the pile, so we get

Thus, the total number of logs in the pile is 6.
Answer:
<h2>
<em>3</em><em>9</em><em> </em><em>units</em></h2>
<em>Solution,</em>
<em>Total </em><em>area=</em><em>Area </em><em>of </em><em>ABCD+</em><em> </em><em>Area </em><em>of </em><em>triangle </em><em>ADE</em>
<em>
</em>
<em>hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>