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4 years ago

-1 2/3 + 3 1/2 as a mixed number

Mathematics

2 answers:

balu736 [363]4 years ago

8 0

1 83/100 because 3-1=2

Harrizon [31]4 years ago

7 0

3 1/6
-1 2/3 + 3 1/2
You add the base numbers first, -1 + 3 to get 2.
Then you add the fraction part of the number: 2/3 + 1/2
You find a common denominator, it's 6.
How do you get 2/3 with a denominator of 6? You multiply 3x2, to get six. what you do to the bottom of the fraction you do to the top. So you multiply both parts by 2.
2x2/3x2= 4/6
You do the same for the other fraction:How do you get the denominator to 6? You multiply by 3.
1x3/2x3= 3/6
Then you add the two new fractions together:
4/6+3/6=7/6 which is simplified into a mixed number as 1 1/6
Since we have another base number you add it to the previous base number we got when we added the mixed numbers. -1 and 3 to get 2. So you add the old base number of 2, to the new base number of 1 to get 3. Then you just put the fraction next to it to get your answer:
3 1/6
Hope it helps.

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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,

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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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Let the distance of Judy from intersection is x and distance of Jackie from intersection is y.

We convert the given information to equations, step by step.

Point 1: When Jackie is 1 mile farther from the intersection than Judy.

This means y is 1 mile more than x.

So,

y = 1 + x

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Distance between is x+ y.

So, x+y is 2 miles more than y.

x+y = y + 2
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x = 2

From point 1 we have:

y = 1 + x = 1+ 2 = 3

So,

Distance of Judy from intersection is 2 miles and distance of Jackie from intersection is 3 miles.

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