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3 years ago

6

Line y = 3x - 1 is transformed by a dilation with a scale factor of 2 and centered at (3, 8). The line’s image is

1 answer:

kondor19780726 [428]3 years ago

5 0

C. Fill in the point (3,8) into the equation. (8)=3(3)-1. 8=9-1. 8=8

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What is the area of a rectangle with a width of 4 2/3 meters and a length of 2 1/6 meters?

inessss [21]

Answer:

Area = 10 1/9 meters squared

Step-by-step explanation:

mixed number to improper fraction:

4 2/3 meters = 14/3 meters

2 1/6 meters = 13/6 meters

Area = length times width:

Area = (14/3) x (13/6)

= 182/18

Simplify fraction:

182/18 = 91/9

improper fraction to mixed number:

91/9 = 10 1/9 meters

AREA = 10 1/9 meters

5 0

3 years ago

The diagram represents a reduction of a triangle by using a scale factor of 0.8.

Alekssandra [29.7K]

Answer:

4.8 inches

Step-by-step explanation:

<em>See comment for complete question</em>

Represent the larger triangle with 1 and the smaller with 2.

So, we have:

$H_1 =6\ in$ -- height of 1

Required

Determine H2 --- Height of 2

To do this we apply dilation formula.

$Original * Scale\ Factor = New$

In this case:

$H_1 * Scale\ Factor = H_2$

Substitute 6 for H1 and 0.8 for Scale Factor

$6* 0.8 = H_2$

$4.8= H_2$

Hence, the height of the smaller triangle is 4.8 inches

3 0

3 years ago

What is the result when this equation is solved for y?

earnstyle [38]

Answer:

Ans:C

Step-by-step explanation:

4x^2+10x-2=6

4x^2+10x=6+2y

4x^2+10x-6=2y

(4x^2+10x-6)÷2=y

y=2x^2+5-3

Hope this is useful :)

3 0

3 years ago

Alexxx [7]

9514 1404 393

Answer:

14.9 cm

Step-by-step explanation:

To find c using the Law of Sines, you must know angle C. That is found from ...

C = 180° -A -B = 180° -150° -12° = 18°

Then the law of sines tells you ...

c/sin(C) = b/sin(B)

c = b·sin(C)/sin(B) = (10 cm)·sin(18°)/sin(12°)

c ≈ 14.9 cm

5 0

3 years ago

Find the axis of symmetry and the vertex of the graph of y = 3x2 + 2x.

hichkok12 [17]

Step-by-step explanation:

<u>Finding Vertex</u>

Given

$y\:=\:3x^2\:+\:2x$

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

The Parabola params are:

$a=3,\:b=2,\:c=0$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{2}{2\cdot \:3}$

$x_v=-\frac{1}{3}$

$\mathrm{Plug\:in}\:\:x_v=-\frac{1}{3}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=3\left(-\frac{1}{3}\right)^2+2\left(-\frac{1}{3}\right)$

$=3\left(-\frac{1}{3}\right)^2-2\cdot \frac{1}{3}$

$=\frac{1}{3}-\frac{2}{3}$ ∵ $3\left(-\frac{1}{3}\right)^2=\frac{1}{3}$

$=\frac{1-2}{3}$

$=\frac{-1}{3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{1}{3}$

$y_v=-\frac{1}{3}$

Therefore the parabola vertex is:

$\left(-\frac{1}{3},\:-\frac{1}{3}\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}$

$a=3$

$\mathrm{Minimum}\space\left(-\frac{1}{3},\:-\frac{1}{3}\right)$

<u>Finding symmetry</u>

For a parabola in standard form $y=ax^2+bx+c$

the axis of symmetry is the vertical line that goes through the vertex $x=\frac{-b}{2a}$

$\mathrm{Axis\:of\:Symmetry\:for}\:y=ax^2+bx+c\:\mathrm{is}\:x=\frac{-b}{2a}$

$a=3,\:b=2$

$x=\frac{-2}{2\cdot \:3}$

$x=-\frac{1}{3}$

Therefore,

$\mathrm{Axis\:of\:Symmetry\:for}\:y=3x^2+2x:\quad x=-\frac{1}{3}$

3 0

3 years ago