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Firlakuza [10]
3 years ago
5

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.

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Pie

Answer:

x = \frac{d+c}{a - b}

Step-by-step explanation:

Equation: ax – c = bx + d

Jalil says it is not possible to isolate x because  x has a different unknown coefficient.

Victoria t. Victoria believes there is a solution

Solving the equation:

ax-c = bx + d

ax - bx= d+c

(a - b)x = d+c

x = \frac{d+c}{a - b}

So, this shows x can be isolated .

Victoria was right .

It was not possible to isolate x if the coefficients of x would be same .

But in the given equation the coefficients of x are not same .

So, Victoria is right.

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3 years ago
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Marilyn bought 6 cans of tuna at $1.29 a can. How much did she spend
nika2105 [10]
1.29*6=7.74 so she spend $7.74
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What is the slope of the line that passes through the points (4, 9) and (1, 6)?
Aleksandr [31]

a = ( 1 , 6 )

b = ( 4 , 9 )

slope =  \frac{y(b) - y(a)}{x(b) - x(a)} \\

Now just need to put the coordinates in the above equation :

slope =  \frac{9 - 6}{4 - 1} \\

slope =  \frac{3}{3} \\

slope = 1

And we're done...♥️♥️♥️♥️♥️

5 0
3 years ago
PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi
Irina18 [472]

Answer:

Question 1: P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}

Question 2:

A. P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}

B. P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}

C. P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}

Step-by-step explanation:

Conditional probability is defined by

P(A|B)= \frac{P(A and B)}{P(B)}

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} }  = \frac{1}{2}

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

P ( Z)= \frac{ 1 }{ 16 }

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

Z: Championships won by the 13 - 15 years old, beeing

P ( Z)= \frac{ 1 }{ 16 }

then

P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

so

P((YorZ) and B)= \frac{3}{16}

By definition,

P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}

3 0
3 years ago
An item costs $440 before tax, and the sales tax is $30.80.
JulsSmile [24]

Answer:

7%

Step-by-step explanation:

30.80%=0.308

0.308/440=0.0007%

0.0007 is the same as 7%

4 0
3 years ago
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