You need to know the value of X to determine whether or not the equation is true.
Answer:
Step-by-step explanation:
<u>Lets verify with Pythagorean:</u>
- 17² = 289
- 13² + 14² = 169 + 196 = 365
- 289 < 365
The angle opposite to a greater side is less than 90° and the sum of the squares are close.
It means all three angles<u> are less than 90°</u>.
With this the triangle is <u>acute</u>.
Correct choice is A.
First find the slope
the slope of the line that passes through (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)
given
(-4,3) and (6,-2)
the slope is (-2-3)/(6-(-4))=(-5)/(6+4)=-5/10=-1/2
so
the equation of a line tat passes through the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)
use (-4,3)
and we found slope=-1/2
y-3=-1/2(x-(-4))
y-3=-1/2(x+4)
if yo want slope intercept form
y-3=-1/2x-2
y=-1/2x+1
if yo want standard form
1/2x+y=1
x+2y=2
Answer:
3384
Step-by-step explanation:
Answer:β=√10 or 3.16 (rounded to 2 decimal places)
Step-by-step explanation:
To find the value of β :
- we will differentiate the y(x) equation twice to get a second order differential equation.
- We compare our second order differential equation with the Second order differential equation specified in the problem to get the value of β
y(x)=c1cosβx+c2sinβx
we use the derivative of a sum rule to differentiate since we have an addition sign in our equation.
Also when differentiating Cosβx and Sinβx we should note that this involves function of a function. so we will differentiate βx in each case and multiply with the differential of c1cosx and c2sinx respectively.
lastly the differential of sinx= cosx and for cosx = -sinx.
Knowing all these we can proceed to solving the problem.
y=c1cosβx+c2sinβx
y'= β×c1×-sinβx+β×c2×cosβx
y'=-c1βsinβx+c2βcosβx
y''=β×-c1β×cosβx + (β×c2β×-sinβx)
y''= -c1β²cosβx -c2β²sinβx
factorize -β²
y''= -β²(c1cosβx +c2sinβx)
y(x)=c1cosβx+c2sinβx
therefore y'' = -β²y
y''+β²y=0
now we compare this with the second order D.E provided in the question
y''+10y=0
this means that β²y=10y
β²=10
B=√10 or 3.16(2 d.p)
