Answer:
Step-by-step explanation:
The constant term of x^2 + 13x – 48 factors into either (3)(-16) or (-3)(16).
Note how 16 - 3 = 13, which is the coefficient of the middle term. Thus, the factors are
(x + 16)(x - 3) which is equivalent to x^2 + 16x - 3x - 48, or x^2 + 13x - 48.
There both types of numbers but a whole number is a number with no exponents or decimals, but decimals are like a broken un half whole number
Hey there,
So let's start off with one part of this equation
. . . 
would be 0.01.
So, we have 0.01 has our answer (so far).
Then we would do 7.1 × 0.01, this would give us
. . .
0.071.Your correct and final answer would be
Hope this helps you!
~Jurgen
Answer:
y2 = C1xe^(4x)
Step-by-step explanation:
Given that y1 = e^(4x) is a solution to the differential equation
y'' - 8y' + 16y = 0
We want to find the second solution y2 of the equation using the method of reduction of order.
Let
y2 = uy1
Because y2 is a solution to the differential equation, it satisfies
y2'' - 8y2' + 16y2 = 0
y2 = ue^(4x)
y2' = u'e^(4x) + 4ue^(4x)
y2'' = u''e^(4x) + 4u'e^(4x) + 4u'e^(4x) + 16ue^(4x)
= u''e^(4x) + 8u'e^(4x) + 16ue^(4x)
Using these,
y2'' - 8y2' + 16y2 =
[u''e^(4x) + 8u'e^(4x) + 16ue^(4x)] - 8[u'e^(4x) + 4ue^(4x)] + 16ue^(4x) = 0
u''e^(4x) = 0
Let w = u', then w' = u''
w'e^(4x) = 0
w' = 0
Integrating this, we have
w = C1
But w = u'
u' = C1
Integrating again, we have
u = C1x
But y2 = ue^(4x)
y2 = C1xe^(4x)
And this is the second solution
Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
Ksju [112]
Well, here's one way to do it at least...
<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>
<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>
<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>
<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>
<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>
<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>
<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>
<span>Now apply the Law of Sines to find the other two angles. </span>
<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>
<span>Two down, one to go... </span>
<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>
<span>So your three angles are: </span>
<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
