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Sunny_sXe [5.5K]
3 years ago
14

Simplify the expression completely. −4g+8−5+ 2g

Mathematics
1 answer:
BartSMP [9]3 years ago
5 0

Answer:

g = 3/2

Step-by-step explanation:

−4g+8−5+ 2g

−4g+ 2g +8-5

-2g + 8 -5

-2g +3

2g = 3

g = 3/2

g = 1 1/2

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The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

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3 years ago
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3 years ago
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prove that the quadrilateral whose vertices are the points A(-1,1), B(-3,4), C(1,5) and D(3,2) is a parallelogram.
Vsevolod [243]

Answer:

Since \overrightarrow{AB} = \overrightarrow{DC} and \overrightarrow{AD} = \overrightarrow{BC}, then the quadrilateral ABCD is a parallelogram.

Step-by-step explanation:

First, we label each point of the quadrilateral with the help of a graphing tool. If the quadrilateral ABCD is a parallelogram, then \overrightarrow{AB} = \overrightarrow{DC} and \overrightarrow{AD} = \overrightarrow{BC}. If we know that A(x,y) =(-1, 1), B(x,y) =(-3,4), C(x,y) = (1,5) and D(x,y) = (3,2), then the measure of each vector is, respectively:

\overrightarrow{AB} = (-3,4)-(-1,1)

\overrightarrow{AB} = (-2, 3)

\overrightarrow{DC} = (1,5)-(3,2)

\overrightarrow{DC} = ( -2,3)

\overrightarrow{AD} = (3,2)-(-1,1)

\overrightarrow{AD} = (4, 1)

\overrightarrow{BC} = (1,5)-(-3,4)

\overrightarrow{BC} = (4, 1)

Since \overrightarrow{AB} = \overrightarrow{DC} and \overrightarrow{AD} = \overrightarrow{BC}, then the quadrilateral ABCD is a parallelogram.

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3 years ago
Factor this question completely
ratelena [41]

Answer:

3b(3a+7c) i used mathpapa.com

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4 years ago
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