Answer:
1.) supplementary m<1+m<2=180 and m<2+m<3=180
2.) congruent, supplementary
Step-by-step explanation:
the answers are in this order above
Yes you did it correctly but change problem 5’s denominator on the right to a 6
Answer:
666.67π .ft^3
<em>here's</em><em> your</em><em> solution</em>
<em>=</em><em>></em><em> </em><em>radius</em><em> of</em><em> </em><em>cone </em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>ft</em>
<em>=</em><em>></em><em> </em><em>height</em><em> of</em><em> </em><em>cone </em><em>=</em><em> </em><em>2</em><em>0</em><em> </em><em>ft</em>
<em>=</em><em>></em><em> </em><em>volume</em><em> of</em><em> </em><em>cone </em><em>=</em><em> </em><em>πr^</em><em>2</em><em>h</em><em>/</em><em>3</em>
<em>=</em><em>></em><em> </em><em>putting</em><em> the</em><em> value</em><em> of</em><em> </em><em>radius</em><em> and</em><em> height</em><em> </em>
<em>=</em><em>></em><em> </em><em> </em><em> </em><em>volume</em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em>^</em><em>2</em><em>*</em><em>2</em><em>0</em><em>/</em><em>3</em><em>π</em>
<em>=</em><em>></em><em> </em><em>volume</em><em> </em><em>=</em><em> </em><em>6</em><em>6</em><em>6</em><em>.</em><em>6</em><em>7</em><em>π</em><em> </em><em>.</em><em>ft^</em><em>3</em>
<em>hope</em><em> it</em><em> helps</em>
The answer for the above mentioned problem is JL = 12.5
Step by step explanation:
Given:
JM = 8
KM = 6
To Find:
JL = ?
Formula to be used:
= JM x ML
In order to find " JL" we must first find "ML",
= JM x ML
= 8 x ML
36 = 8 x ML
36/8 = ML
ML = 4.5
Now JL = 4.5+8
= 12.5
Thus the value of JL = 12.5
Answer:
(A)
with
.
(B)
with 
(C)
with 
(D)
with
,
Step-by-step explanation
(A) We can see this as separation of variables or just a linear ODE of first grade, then
. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form
with
real.
(B) Proceeding and the previous item, we obtain
. Which is not a vector space with the usual operations (this is because
), in other words, if you sum two solutions you don't obtain a solution.
(C) This is a linear ODE of second grade, then if we set
and we obtain the characteristic equation
and then the general solution is
with
, and as in the first items the set of solutions form a vector space.
(D) Using C, let be
we obtain that it must satisfies
and then the general solution is
with
, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).