Multiply q by r, subtract p from the result, then triple what you have

Which could you use to prove triangle AED is congruent to triangle CEB?

Hunter-Best [27]

Answer:

Option (C)

Step-by-step explanation:

Given:

In right triangles ΔAED and CEB,

m∠AED = m∠CEB = 90°

DE ≅ BE

AD ≅ BC

To prove:

ΔAED ≅ ΔCEB

Statements Reasons

1). m∠AED = m∠BC = 90° 1). Given

2). DE = BE 2). Given

3). AD = BC 3). Given

4). ΔAED ≅ ΔCEB 4). By HL theorem of congruence

Option (C) is the answer.

4 0

2 years ago

The rectangle shown has a perimeter of 154 cm and the given area. Its length is 5 more than five times its width. Write and solv

Andreyy89

Answer:

2(L+W) = 34

L + W = 17

.

L*W = 52

.

L = 2W+5

.

2L + 2W = 34

L - 2W = 5

--------------- add

3L = 39

L = 13

8 0

2 years ago

Read 2 more answers

The ratio of the measures of the sides of a triangle is 9:15:5. If the perimeter of the triangle is 130 feet, find the measures

Alex Ar [27]

5x + 9x +15x = 130
29x = 130
x = 4.4827586207 
So the triangle sides =
 22.4137931034 
 40.3448275862 
 67.2413793103

5 0

3 years ago

Joseph wants to estimate 1 3/7 . 5 4/5 To estimate, he simplifies the expression 1 1/2 . 6 Is Joseph’s estimate an underestimate

Zolol [24]

Well if her simplifies 1 3/7 to 1 1/2 and 5 4/5 to 6 then he would be overestimating because 1 1/2 is greater than 1 3/7 and 6 is greater than 5 4/5

Hope this helps!

7 0

2 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

2 years ago