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slamgirl [31]
3 years ago
12

Someone help please, please be correct.

Mathematics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

Which question?

Step-by-step explanation:

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Use the drop-down menus below to state the sequence of transformations that maps Figure W onto Figure X in the animation below.
iragen [17]

Answer:

Dilate, Reflect

Not congruent, dilations are used

Step-by-step explanation:

Dilate the figure by a scale factor of 1/3

Reflecting the resulting shape will map the figure

Step 1: Dilate by 1/3

Step 2: Reflect over the y-axis

Since dilations are used, the figures are not the same, as it changes area and side lengths.

-Chetan K

3 0
2 years ago
Isolated triangle 68 perimeter equal sides more than 10 ft use x for the shorter side formula
Ugo [173]

Answer:

Step-by-step explanation:

Find the perimeter of an isosceles triangle whose equal sides have a size of 10 m each and the angle between them equal to 30°. We need to know all sides in order to find the perimeter of this triangle. Let x be the base of this isosceles triangle.

6 0
2 years ago
7divided by 43.4 pls help
Vikki [24]

0.161      (they said explain it so)                                                                                                                                                                                                                                                                              

8 0
3 years ago
Consider the following system of two linear equations:
harkovskaia [24]

Hello,

Answer:

The last graph (2;3)

Step-by-step explanation:

2x + 3y = 12

2x – 3y = 0

2x + 3y = 12

2x = 3y

3y + 3y = 12 ⇔ 6y = 12 ⇔ y = 12/6 = 2

2x + 3 × 2 = 12 ⇔ 2x + 6 = 12 ⇔ 2x = 6 ⇔ x = 6/2 = 3

4 0
2 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}

#CarryOnLearning

#BrainlyMathKnower

#5-MinutesAnswer

7 0
2 years ago
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