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olga_2 [115]
3 years ago
8

An isosceles triangle has two sides of equal length, a, and a base, b. The perimeter of the triangle is 15.7 inches, so the equa

tion to solve is 2a + b = 15.7.
Which lengths make sense for possible values of b? Check all that apply.

–2 in.
0 in.
0.5 in.
2 in.
7.9 in.
Mathematics
2 answers:
puteri [66]3 years ago
8 0

Answer:

He's right the answers is 0.5 and and 2

Step-by-step explanation:

C is 0.5 in and D 2 in

Orlov [11]3 years ago
4 0

it cant be a negative number or zero

 so now you are left with 0.5, 2 and 7.9

15.7 - 0.5 = 15.2 ( 15.2/2 = 7.6)  this is possible

15.7 - 2 = 13.7 ( 13.7/2 = 6.85) this is possible

15.7 - 7.9 = 7.8 ( 7.8/2 = 3.9) (7.9/2 = 3.95) since the sides would be shorter than half the base this is not possible to form a triangle


so b can be either 0.5 or 2



Please mark brainliest

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LekaFEV [45]

The correct answer for the polar coordinates will be C, A, E, B, D

<h3>What is a polar coordinate system?</h3>

The polar coordinate system is a two-dimensional coordinate system in which a distance from a reference point and an angle from a reference direction identify each point on a plane. The pole is the reference point, and the polar axis is the ray from the pole in the reference direction.

The conversion relations between rectangular and polar coordinates can be used to find the polar coordinate equivalents. Those relations are ...

x = r·cos(θ)

y = r·sin(θ)

x² +y² = r²

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Using the expression for x, this is ...

r·cos(θ) = 2

r = 2/cos(θ) . . . . divide by cos(θ)

r = 2·sec(θ) . . . . use the trig identity

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From above, this becomes ...

r² = 36

r = 6 . . . . . . take the square root

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From above, this becomes ...

r² = 2·r·sin(θ)

r = 2·sin(θ) . . . . . . divide by r

x=√3y

Using the above relations, this is ...

r·cos(θ) = √3·r·sin(θ)

1/√3 = sin(θ)/cos(θ) . . . . . . divide by √3·r·cos(θ)

tan(θ) = 1/√3   ⇒   θ = arctan(1/√3)

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x=y

From above, this is ...

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1 = sin(θ)/cos(θ) = tan(θ) . . . . divide by r·cos(θ)

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However, the tangent function is periodic with period π, so θ = arctan(k)+π is also equivalent to the rectangular equation. It is the opposite ray, so forms the complete line when joined with the first ray.

To know more about polar coordinate system follow

brainly.com/question/14965899

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olasank [31]

                                             Question # 17

<u><em>Solution:</em></u>

<em>Graphing the vertices of a triangle and transforming  by first translating</em>

<em>(x, y) → (x -3, y - 3) and then reflecting in the line y = x.</em>

As A(-1, 1), B(0, 3) and C(3, 4)

After Translation: (x, y) → (x - 3, y - 3)

A(-1, 1) → A'(-4, -2)

B(0, 3) → B'(-3, 0)

C(3, 4) → C'(0, 1)

After Reflection: in the line y = x

The reflection of the point (x, y) across the  

line y = x is the point (y, x).

A'(-4, -2) → A''(-2, -4)

B'(-3, 0) → B''(0, -3)

C'(0, 1) → C''(1, 0)

Pleas check the top right side of the <em>attached </em><em>figure a</em><em>.</em>

<em>Is the image same if the order of the transformation is switched ? (Reflection, then glide). Changing the order of transformation.</em>

<em>A(-1, 1), B(0, 3) and C(3, 4) </em>

<em> </em><em>After Reflection: in the line y = x </em>

<em>The reflection of the point (x, y) across the  line y = x is the point (y, x). </em>

<em> A(-1, 1) </em>→<em> A'(1, -1) </em>

<em>B(0, 3) </em>→<em> B'(3, 0) </em>

<em>C(3, 4) </em>→<em> C'(4, 3) </em>

<em> </em>

<em>After Translation: (x, y) </em>→<em> (x - 3, y - 3) </em>

<em> A'(1, -1) </em>→<em> A''(-2, -4) </em>

<em>B'(3, 0) </em>→<em> B''(0, -3) </em>

<em>C'(4, 3) </em>→<em> C''(1, 0) </em>

<em>No, the image is </em><em>no longer the same </em><em>if </em><em>order of the transformation</em><em> is </em><em>switched</em><em> as shown at the </em><em>right bottom of figure a</em><em>.</em>

<em />

<em>Answering the YES and NO Questions:</em>

<h3><em>Is the composition a glide reflection?</em></h3>

No, the composition is not a glide reflection as glide reflection is commutative. If we reverse the direction of the composition, the outcome will be affected.

Check the figure at top right side of figure a, and compare it with the figure at the bottom right side of figure a. They are not the same.

<h3><em /></h3><h3><em>Is the image same if the order of the transformation is switched ? (Reflection, then glide)</em></h3>

<em>No, the image is no longer the same if order of the transformation is switched as shown at the right bottom figure of </em><em>figure a</em><em>.</em>

<em>The </em><em>top right side of</em><em> </em><em>figure a</em><em> was transformed after first translation of (x, y) </em>→<em> (x - 3, y - 3) and then doing reflection across the line y = x. While the </em><em>bottom right side of figure a</em><em> was made by first reflection across the line y= x and then translation of (x, y) </em>→<em> (x - 3, y - 3. So, after changing the order both figures do not remain the same. </em>

<em />

                                          Question # 18

<em>Solution:</em>

<em>Graphing the vertices of a triangle and transforming by first 90 rotating clockwise and then reflection in the y axis.</em>

<em>A(-1, 1), B(0, 3) and C(3, 4) </em>

<em>After transforming by first 90 rotating clockwise about the origin. The new position of point (x, y) will be (y, -x) </em>

<em> A(-1, 1) </em>→<em> A'(1, 1) </em>

<em>B(0, 3) </em>→<em> B'(3, 0) </em>

<em>C(3, 4) </em>→<em> C'(4, -3) </em>

<em> </em>

<em>After Reflection: in the line y axis </em>

<em>The reflection of the point (x, y) in the line  line y-axis is the point (-x, y). </em>

<em> A'(1, 1) </em>→<em> A''(-1, 1) </em>

<em>B'(3, 0) </em>→<em> B''(-3, 0) </em>

<em>C'(4, -3) </em>→<em> C''(-4, -3) </em>

<em>Please check the </em><em>top right side</em><em> of </em><em>figure b.</em>

<em>If the order of the transformation were performed in reverse order ? (Reflection, then rotation), would the final image by the same?</em>

A(-1, 1), B(0, 3) and C(3, 4)

After Reflection: in the line y-axis

The reflection of the point (x, y) in the line y-axis is the point (-x, y).

A(-1, 1) → A'(1, 1)

B(0, 3) → B'(0, 3)

C(3, 4) → C'(-3, 4)

After transforming by first 90 rotating clockwise about the origin. The new position of point (x, y) will be (y, -x)

A'(1, 1) → A''(1, -1)

B'(0, 3) → B''(3, 0)

C'(-3, 4) → C''(4, 3)

<em>No, the image is </em><em>no longer the same </em><em>if </em><em>order of the transformation</em><em> is </em><em>switched</em><em> as shown at the </em><em>right bottom of figure b</em><em>.</em>

<em />

<em>Answering the YES and NO Questions:</em>

<em>If the transformation were performed in reverse order (reflection. then rotation), would the final image be the same?</em>

No, if we reverse the direction of the composition, the final image would not be same.

Check the figure at <em>top right side</em> of <em>figure b</em>, and compare it with the at the<em> bottom right side </em><em>of</em><em> </em>figure b. They are not the same.

<em />

<em>Would the reflection of triangle ABC in the line y = x have produced the same final image?</em>

<em>Yes, the reflection of triangle ABC in the line y = x would have produced the same final image. Compare the figure of the </em><em>bottom left</em><em> with the </em><em>bottom right </em><em>of </em><em>figure b</em><em>.</em>

<em />

<em>As A(-1, 1), B(0, 3) and C(3, 4) </em>

<em> After Reflection: in the line y = x </em>

<em>The reflection of the point (x, y) across the </em>

<em>line y = x is the point (y, x). </em>

<em> A(-1, 1) → A''(1, -1) </em>

<em>B(0, 3) →  B''(3, 0) </em>

<em>C(3, 4) →  C''(4, 3)</em>

<em>Hence, the reflection of triangle ABC in the line y = x would have produced the same final image. Compare the figure of the </em><em>bottom left</em><em> with the </em><em>bottom right </em><em>of </em><em>figure b</em><em>.</em>

<em />

<em>Keywords: transformation, reflection, rotation, translation, triangle</em>

<em>Learn more about different transformations from brainly.com/question/11914738</em>

<em>#learnwithBrainly</em>

3 0
4 years ago
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