Answer:
When two lines are cut by a transversal, the pairs of angles on one side of the transversal and inside the two lines are called the consecutive interior angles . If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary
Step-by-step explanation:
Sanjay, because quarts are twice as big as pints.
Step-by-step explanation:
Given
Milk quantity for Lisa = 3 quarts
Milk quantity for Sanjay = 12 pints
We have to convert both quantities in same unit so that we can compare both quantities.
We know that there are two pints in a quart.
So,
Lisa will have:
3*2 = 6 pints of milk
Hence,
Sanjay, because quarts are twice as big as pints.
Keywords: Units, measurement
Learn more about units at:
#learnwithBrainly
Answer:
<h2>x = 6.5 cm</h2>
Step-by-step explanation:
The segment 20.2 cm is the tangent. Therefore x and 20.2cm are perpendicular.
Use the Pythagorean theorem:

We have

Substitute:
<em>use (a + b)² = a² + 2ab + b²</em>
<em>subtract x² from both sides</em>
<em>subtract 216.09 from both sides</em>
<em>divide both sides by 29.4</em>

1.) X=-2
2.) 2x-y+4=0
Hope this helps! :))
Take the homogeneous part and find the roots to the characteristic equation:

This means the characteristic solution is

.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form

. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With

and

, you're looking for a particular solution of the form

. The functions

satisfy


where

is the Wronskian determinant of the two characteristic solutions.

So you have




So you end up with a solution

but since

is already accounted for in the characteristic solution, the particular solution is then

so that the general solution is