All categories

3 years ago

Given: BD = BF DE ⊥ BC , FK ⊥ AB Prove: ED ≅ FK

Mathematics

2 answers:

sergeinik [125]3 years ago

8 0

This is the picture I am not very good at proofs but hopefully someone can solve it.

Simora [160]3 years ago

6 0

Answer:

Side ED ≅ FK

Step-by-step explanation:

The given parameters are BD = BF, DE ⊥ BC and FK ⊥ AB

We have to prove ED ≅ FK

In the given triangles ΔBDE and ΔBKF

BD = BF

∠B is common and ∠BDE ≅ ∠KFB = 90°

Since these triangles fulfill the property of congruence AAS (One side and two adjacent angles are equal) therefore ΔBDE ≅ ΔBKF

Therefore ED ≅ FK

Hence proved.

You might be interested in

Help please I will give brainleist

Natali5045456 [20]

Answer:

The answer is 80%

8 0

2 years ago

dalvyx [7]

Answer:

38 is the correct solution

3 0

2 years ago

Can someone please help me with 6-9 I'm having a lot of trouble.

USPshnik [31]

If u subtract 6-9 u would get a -3 because the 9 is bigger than the 6 so thats how you get a -3.

3 0

3 years ago

3. (6 Points). Solve the initial value problem y'-y.cosx=0, y(pi/2)=2e

Stells [14]

Answer:

$y=2e^{sin(x)}$

Step-by-step explanation:

Given equation can be re written as

$\frac{\mathrm{d} y}{\mathrm{d} x}-ycos(x)=0\\\frac{\mathrm{d} y}{\mathrm{d} x}=ycos(x)\\\\=> \frac{dy}{y}=cox(x)dx\\\\Integrating \\ \int \frac{dy}{y}=\int cos(x)dx \\\\ln(y)=sin(x)+c$ ............(i)

Now it is given that y(π/2) = 2e

Applying value in (i) we get

ln(2e) = sin(π/2) + c

=> ln(2) + ln(e) = 1+c

=> ln(2) + 1 = 1 + c

=> c = ln(2)

Thus equation (i) becomes

ln(y) = sin(x) + ln(2)

ln(y) - ln(2) = sin(x)

ln(y/2) = sin(x)

$y= 2e^{sinx}$

7 0

3 years ago

The Tennis Club decided to buy T-shirts for its members. Action Sports Apparel quoted the following prices for the T-shirts.

Hatshy [7]

Answer:

C. c(t) = 21 + 12t

Step-by-step explanation:

6 0

2 years ago