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3 years ago

Just look at the picture it doesn’t have words

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

Answer: c 6

Step-by-step explanation:

3 2/3 +2.3...

2.3....=2 1/3

3 2/3 + 2 1/3= 5 3/2

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I only need 12 13 14

Ulleksa [173]

There isn’t any picture you may have to repost

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3 years ago

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Which of the following is not the same as 1.69 meters? 0.0169 km 0.169 dkm 169 cm 1,690 mm

bezimeni [28]

Answer:

0.0169 km.

Step-by-step explanation:

km - kilometers (One kilometer is equal to 1000 meters.)

dkm - dekameters (One dekameter is equal to 10 meters.)

cm - centimeters (One centimeter is equal to 0.01 meters.)

mm - millimeters (One millimeter is equal to 0.1 meters.)

Used a calc to check each of the values to see if they equal 1.69m and km was the one that wasn't equal. Instead it was 16.9.

3 0

3 years ago

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The team ratio of games won to games played was 2 to 7. If the team played 49 games, how many games did the team fail to win?

Nikolay [14]

Answer:

they failed to win 35 games

6 0

3 years ago

Take the points people

Reika [66]

THANK YOU SO MUCHHHH

5 0

3 years ago

Read 2 more answers

The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

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

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)

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

Using rule (i), we get

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

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

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

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

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

Let us assume that point A is A(m,n).

Using rule (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).

4 0

2 years ago