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Mathematics

2 answers:

Zepler [3.9K]3 years ago

7 0

Answer:

They are not congruent would be my guess cause the sides are not the same my bad if it is wrong lil br uh

Step-by-step explanation:

In geometry, two figures or objects are <u><em>congruent</em></u> if they have the same shape and size, or if one has the same shape and size as the mirror image of the other. This means that either object can be repositioned and reflected (but not resized) so as to coincide precisely with the other object.

djyliett [7]3 years ago

3 0

Answer:

They are not congruent because they are not facing the same way.

Step-by-step explanation:

SAS stands for "side, angle, side" and means that we have two triangles where we know two sides and the included angle are equal. If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

That is which postulate proves congruency.

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3. Use substitution to solve the system of

Aleksandr-060686 [28]

Answer:

infinite solutions

Step-by-step explanation:

the lines are coincident, so there are infinite solutions

5 0

3 years ago

The point (5 5/8, 2 1/4) lies on a line that represents a proportional relationship between two units of measure.

neonofarm [45]

9514 1404 393

Answer:

a) k = 2/5

b) y = 2/5x

c) y = 2 3/5

Step-by-step explanation:

The proportional relation is described by the equation ...

y = kx . . . . . . k is the constant of proportionality

The value of k can be found by dividing the equation by x:

y/x = k = (2 1/4)/(5 5/8) = (9/4)/(45/8) = 18/45

k = 2/5

As above, the equation is ...

y = kx

y = 2/5x

Using 6.5 for x, we have ...

y = (2/5)(6.5) = 2.6 = 2 3/5

5 0

3 years ago

|. Identify the following Pōints of each values.Write your ans

Dmitry_Shevchenko [17]

<h2>✒️VALUE</h2>

$\\ \quad \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\ \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\ \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\ \tt \qquad DF = EF \\ \\ \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\ \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\ \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\ \implies\tt10x + 4x = 5 - 61 \\ \\ \implies\tt14x = -56 \\ \\ \implies \red{\boxed{\tt x = -4}}\end{array} \\ \\ \\ \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\ \qquad\textsf{Solving for }y, \\ \\ \qquad\tt PR = QR \\ \\ \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\ \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\ \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\ \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\ \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\ \implies\red{\boxed{ \tt y = 3}} \end{array} \\ \\ \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\ \qquad\textsf{Solving for }x, \\ \\ \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}$