So.. let's say ∡AGB and ∡BGC are say "a" units wide, and ∡CGD and ∡DGE are "b" units wide
notice the picture below
all angles added up together, will make up just a flat line, or 180°
now, notice in the picture, the ∡BGD is really just " a + b " wide, notice the green angle in the picture of ∡BGD, well, we know what a + b is
Answer:
1000 g = 1 kg
Step-by-step explanation:
1000 GRAMS IS EQUAL TO 1 KILOGRAM
1/5 five being the denominator, because in math you must always simplify
Answer:
m∠5 = 44°; m∠7 = 44°
Step-by-step explanation:
Angle 4 and 1 are supplementary (they make up a line) and their sum is equal to 180 degrees.
Subtracting the measure of angle 4 from 180 degrees gives the measure of angle 1. (180 - 136 = 44).
So Angle 1's measure is 44 degrees.
According to the Corresponding Angle Postulate, Angle 1 and Angle 5 are congruent. Therefore, m∠5 = 44°
According to the Vertical Angles Postulate (if two angles are vertical, they are congruent), ∠5 ≅ ∠7, meaning that m∠5 = m∠7.
So m∠7 = 44°
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
