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melomori [17]
2 years ago
13

Point D is in the interior of ZABC, m ABC = 10x - 7, m_ABD = 6x + 5, and m DBC = 36. What is m ABD?

Mathematics
1 answer:
mestny [16]2 years ago
5 0

Answer:

<h3>77</h3>

Step-by-step explanation:

If Point D is in the interior of <ABC, then <ABD+<BDC = <ABC

Given

<ABC = 10x - 7,

<ABD = 6x + 5

<DBC = 36

Substituting the given parameters into the formula above;

ABD+<DBC = <ABC

6x + 5+ 36 = 10x-7

6x+41 = 10x-7

collect like terms

6x-10x = -7-41

-4x = -48

divide both sides by -4

-4x/-4 = -48/-4

x = 12

To get <ABD, we will substitute x = 12 into the expression <ABD = 6x + 5

<ABD = 6(12)+5

<ABD = 72+5

<ABD =77

Hence the value of <ABD is 77

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Answer:

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Step-by-step explanation:

8 0
2 years ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

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

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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

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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

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Answer:

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Step-by-step explanation:

To differentiate h(x)=f(x)g(x) we will need the product rule:

h'(x)=f'(x)g(x)+f(x)g'(x).

We have h'(x)=f(x)g'(x), so the following equation is true by the transitive property:

f'(x)g(x)+f(x)g'(x)=f(x)g'(x)

By subtraction property we have:

f'(x)g(x)=0

Since g(x) \neq 0, then we can divide both sides by g(x):

f'(x)=\frac{0}{g(x)}

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This implies f(x) is constant.

So we have that f(x)=c where c is a real number.

Since f(0)=1 and f(0)=c, then by transitive property 1=c.

So f(x)=1.

Checking:

h(x)=1 \cdot g(x)

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So the following conditions were met.

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