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2 years ago

Pleaseeeee need this quickly 50 points!

Mathematics

2 answers:

timofeeve [1]2 years ago

8 0

Answer:

words are a little blurry but I think it is c|

Step-by-step explanation:

It is actually 50 points for us because it takes away half of what you put in that's why it is 50 for us

aivan3 [116]2 years ago

4 0

its CL i think im very sorry if im wrong

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2 years ago

0.139 times what number could get me to 100? PLEASE ANSWER, FAST

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Answer:

719.4244604

Step-by-step explanation:

100/0.139=719.4244604

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3 years ago

How much longer is the hypotenuse of the triangle than its shorter leg?

RUDIKE [14]

This is an incomplete question, here is a complete question and image is also attached below.

How much longer is the hypotenuse of the triangle than its shorter leg?

a. 2 ft

b. 4 ft

c. 8 ft

d. 10 ft

Answer : The correct option is, (b) 4 ft

Step-by-step explanation:

Using Pythagoras theorem in ΔACB :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(AB)^2=(AC)^2+(BC)^2$

Given:

Side AC = 6 ft

Side BC = 8 ft

Now put all the values in the above expression, we get the value of side AB.

$(AB)^2=(6)^2+(8)^2$

$AB=\sqrt{(6)^2+(8)^2}$

$AB=10ft$

Now we have to calculate the how much longer is the hypotenuse of the triangle than its shorter leg.

Difference = Side AB - Side AC

Difference = 10 ft - 6 ft

Difference = 4 ft

Therefore, the 4 ft longer is the hypotenuse of the triangle than its shorter leg.

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3 years ago

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Identify the three similar right triangles in the given diagram.

irga5000 [103]

Answer:

B. ΔABD, ΔADC, ΔDBC

Step-by-step explanation

Step -1 In ΔABD and ΔADC (from figure).

∠DAB=∠CAD (common in both triangles) ,

∠DBA=∠CDA =90 degree, and

∠BDA=∠DCA (rest angle of the two triangles).

therefore ΔABD similar to ΔADC (by AAA similarity theorem).

Step -2 In ΔDBC and ΔADC (from figure).

∠DCB=∠ACD (common in both triangles) ,

∠DBC=∠ADC =90 degree, and

∠CDB=∠CAD (rest angle of the two triangles).

therefore ΔDBC similar to ΔADC (by AAA similarity theorem).

Step -3 In ΔABD and ΔDBC (from figure).

∠BDA=∠BCD (because , ∠ACD=ADB from stap-1 and ∠ACD=∠BCD from figure) ,

∠DBA=∠CBD =90 degree, and

∠BAC=∠BDC (rest angle of the two triangles).

therefore ΔABD similar to ΔDBC (by AAA similarity theorem).

In the above step- ΔABD similar to ΔADC, ΔDBC similar to ΔADC and ΔABD similar to ΔDBC.

Hence ΔABD, ΔADC, ΔDBC similar to each other in the given figure.

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