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2 years ago

Angle ABC measures 140°. Angle DBC bisects ∠ABC. Angle EBC bisects ∠DBC. What is the measure of ∠EBC?

Mathematics

1 answer:

Brrunno [24]2 years ago

5 0

Answer:

$\angle{EBC}=35\textdegree$

An angle bisector divides the angle measure into two equal parts. In triangle ABC, B is the point at which the angle lies.

Step-by-step explanation:

Picture segment BD drawn from point B of tri.ABC where it connects to segment AC, creating point D. Now picture E drawn on the newly created segment DC where it connects to point B and down to point C. Here's a drawing to better show this:

It's bisecting the bisected angle ABC. So, 140/2 = 70, and 70/2 = 35.

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Solve to find the values of a and b.<br>4a + 7b = 38<br>9a + 7b = 68

Nataliya [291]

4a + 7b = 38 -> (1)
9a + 7b = 68 -> (2)

Subtract (1) from (2),
9a - 4a + 7b - 7b = 68 - 38
5a = 30
a = 6

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3 years ago

Four friends made 55 cupcakes and shared them equally.how many cupcakes did each friend get? PLZ HELP ME:(

Drupady [299]

Answer:

13 3/4 cupcakes for each person

Step-by-step explanation:

55 cupcakes and 4 friends

55/4 = 13 3/4

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2 years ago

Read 2 more answers

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Leno4ka [110]

My guess would be A ! Hope this helps

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3 years ago

Consider the sequence 6, 11, 16, 21 . . . . What is the 50th term in this sequence?

Damm [24]

Answer:

251

Step-by-step explanation:

Tn=5n+1

e.g.T1=5(1)+1

e.g.T2=5(2)+1

=11

e.g.T3=5(3)+1

=16

e.g.T4=5(4)+1

=21

Answer:T50=5(50)+1

=251

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3 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago