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2 years ago

What is the vaule of x

Mathematics

1 answer:

blondinia [14]2 years ago

4 0

Answer:

Step-by-step explanation:

Look at the picture.

If ABC is any triangle and AD bisects (cuts in half) the angle BAC, then

$\dfrac{AB}{DB}=\dfrac{AC}{DC}$

In our triangle we have the proportion:

$\dfrac{AD}{AC}=\dfrac{BD}{BC}$

We have

AD = 2.25, AC = 3, BC = 4, BD = x.

Substitute:

$\dfrac{2.25}{3}=\dfrac{x}{4}$

$\dfrac{x}{4}=0.75$ multiply both sides by 4

$\boxed{x=3}$

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NEED ANSWER ASAP!! TYYY SO MUCH IN ADVANCE

Kipish [7]

Answer:

24.84

Step-by-step explanation:

use Pythagoras theorem

h=√p²+b² = √(19)²+(16)² = √617 =24.84

4 0

2 years ago

WHY IS THIS GETTING DELETED?? I NEED HELP

aliina [53]

If this exact question is repeatedly deleted, it's probably because of the ambiguity of the given equation. I see two likely interpretations, for instance:

$\dfrac{(5\times5)^k}{5^{-8}} = 5^3$

$\dfrac{5\times 5^k}{5^{-8}} = 5^3$

If the first one is what you intended, then

$\dfrac{(5\times5)^k}{5^{-8}} = \dfrac{(5^2)^k}{5^{-8}} = \dfrac{5^{2k}}{5^{-8}} = 5^{2k-(-8)} = 5^{2k+8} = 5^3$

and it follows that

2k + 8 = 3 ==> 2k = -5 ==> k = -5/2

If you meant the second one, then

$\dfrac{5\times 5^k}{5^{-8}} = \dfrac{5^1\times5^k}{5^{-8}} = \dfrac{5^{k+1}}{5^{-8}} = 5^{k+1-(-8)} = 5^{k+9} = 5^3$

which would give

k + 9 = 3 ==> k = -6

And for all I know, you might have meant some other alternative... When you can, you should include a picture of your problem.

3 0

2 years ago

Can someone help me with these 2 questions please and thank you?

Inessa05 [86]

2 × 3 = 6. 2. 7×3 =21

4 0

3 years ago

What is the length of diagonal BD in rectangle ABCD if BC=5 and CD=8.

svetlana [45]

I hope this helps you

6 0

2 years ago

8.54 as a place value

alexandr402 [8]

Eight and fifty four hundredths

8 . 5 4
Ones Tenths Hundredths

7 0

2 years ago