All categories

2 years ago

Can you use the ASA postulate or the AAS theorem to prove the triangles congruent?

Mathematics

2 answers:

Molodets [167]2 years ago

8 0

Answer:

Step-by-step explanation:

Let us name the given figure, there are two triangles that areΔABO and ΔCOD.

Thus, from ΔABO and ΔCOD, we have

∠ABO=∠DCO(Given in the question)

BO=DO( given in question)

∠AOB=∠COD(Vertically opposite angles)

Therefore, by ASA rule,

ΔABO ≅ΔCOD

Therefore, only ASA rule can be applied on the given figure.

Hence, option C is correct.

Naya [18.7K]2 years ago

3 0

Angle side angle theorem i believe?
angle (the one with the arc), the tick (side), then the angle (the vertical angle).

You might be interested in

Is relation to t a function? is the inverse of relation t a function?

Alona [7]

The relation t is one-to-one, so it is a function and so is its inverse

3 0

2 years ago

The reflection of point S across the y-axis is point

Ostrovityanka [42]

Point V
Also point V
(-5,-5.5)

3 0

3 years ago

Find the area of a parallelogram with base (b) and height (h)

anyanavicka [17]

Answer is C) 1,095.2 cm2
Formula for area for parallelogram is = b.h

7 0

3 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

3x- 5=1 what does x represent

V125BC [204]

Answer:

x=2

Step-by-step explanation:

3x=5+1

3x=6

x=6/3

x=2

8 0

2 years ago

Read 2 more answers