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2 years ago

Which congruence theorems can be used to prove ΔABR ≅ ΔACR? Select three options

Mathematics

2 answers:

Mashcka [7]2 years ago

8 0

Answer:

SAS , ASA, SSS

Step-by-step explanation:

SAS , side A= side A . Angle B= Angle C, side R= side R

ASA, Angle A= Angle A . Side B= side C, Angle R= Angle R

SSS

side A= side A . Side B= side c, side R= side R

Leokris [45]2 years ago

7 0

Answer:

<h2>They are congruent by SAS postulate.</h2>

Step-by-step explanation:

Choice for this question are

In the image attached, you can observe the pairs of triangles we are talking about. Notice that they are right triangles, which means the congruence postulates we could use are Hypothenuse-Leg, Leg-Leg, Leg-Angle and Hypothenuse Angle, however, this choices are not useful.

We know by given that

$AC \cong AB$ and $RC \cong RB$

We also know that $\angle B \cong \angle C$

So, we have two corresponding sides congruent, and the pair of corresponding angle formed by those sides are also congruent.

Therefore, by Side-Angle-Side

$\triangle ABR \cong \triangle ACR$

So, the right answer is SAS.

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I’ll fail if I don’t get this right pls help

son4ous [18]

Step-by-step explanation:

In order to find the number of solutions, we have to expand both sides out first.

$3(x - 2) + 7x = \frac{1}{2} (6x - 2)$

First, expand 3(x-2), which will obtain 3x-6.

Next, expand (1/2)(6x-2), which gives 3x - 1.

The result at this point should be as shown below:

$3x - 6 + 7x = 3x - 1$

Now, we would shift all x terms to the left and all number terms to the right.

$3x + 7x - 3x = - 1 + 6$

After simplifying, the final equation would be:

$7x = 5$

This is a linear equation and it has one solution.

$x = \frac{5}{7}$

7 0

2 years ago

Read 2 more answers

Is ( 0,0) a solution to this system? y<=x^2-4 y>2x-1?

Xelga [282]

Answer:

The point $(0,0)$ is not a solution of the system of inequalities

Step-by-step explanation:

we have

$y\leq x^{2} -4$ -----> inequality A

$y> 2x-1$ -----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities

then

the ordered pair must be satisfy the inequalities of the system

Verify

For $x=0, y=0$

substitute the value of x and the value of y in the inequalkity A and in the inequality B

Inequality A

$0\leq 0^{2} -4$

$0\leq -4$ -------> is not true

therefore

The point $(0,0)$ is not a solution of the system of inequalities

3 0

2 years ago

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Using the graph below, select all statements that are true.

Harrizon [31]

Answer:

b and E and c

Step-by-step explanation:

A is false because some elements have the same image like 2.3 and 2.2 , 2.5 ...
B is true because the image of an integer remains the same using the greatest integer function. So C is true
E is true

6 0

2 years ago

PLS HELP!!! WILL GIVE 100 points

12345 [234]

Answer:

$125.00

Step-by-step explanation:

$185.50 x 2 + $4.00 =

$371 + $4.00 =

$375 =

so $375 ÷ 3 = $125.00

So Rayne sell each desk for $125.00

3 0

2 years ago

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Simplify this problem

ikadub [295]

Answer:

$\boxed{ \frac{ \sqrt[3]{ {x}^{11} } }{4} }$

Step-by-step explanation:

$= > \frac{ {x}^{4} }{ \sqrt[3]{64x} } \\ \\ = > \frac{ {x}^{4} }{ {(64x)}^{ \frac{1}{3} } } \\ \\ = > \frac{ {x}^{4} }{ ({64}^{ \frac{1}{3} } )\times ({x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{ ({( {4}^{3} )}^{ \frac{1}{3} }) \times( {x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{ ({4}^{ \cancel{3} \times \frac{1}{ \cancel{3}} } ) \times( {x}^{ \frac{1}{3} } )} \\ \\ = > \frac{ {x}^{4} }{4 {x}^{ \frac{1}{3} } } \\ \\ = > \frac{ {x}^{4 - \frac{1}{3} } }{4} \\ \\ = > \frac{ {x}^{ \frac{12 - 1}{3} } }{4} \\ \\ = > \frac{ {x}^{ \frac{11}{3} } }{4} \\ \\ = > \frac{ \sqrt[3]{ {x}^{11} } }{4}$

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2 years ago