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faust18 [17]
3 years ago
14

How do you solve this?

Mathematics
1 answer:
notka56 [123]3 years ago
8 0
1. Multiply 10 and 1, then add 9. Then the fraction will be 19/10.
2. Multiply 19/10 and (-5/9). That will give you -95/90, which reduces to -19/18
3. Add -19/18 and 7/3. 
4. The answer is 23/18 
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Expression A is 18 adding .5 and expression B is 18.5 multiplying buy 2.
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HELP
Zanzabum
By looking at the formula you can see that the quadratic part is 0 if you fill in x=3 (you know, 3-3 = 0). The outcome of the formula then becomes y= 0+5.

That gives us point (3,5) as the top of the parabola. So there you go, answer A.
4 0
3 years ago
Distance on graph
Mnenie [13.5K]
The graph can be divided into two shapes. A rectangle and a right triangle.

The distance that the rover will cover if it completes one circuit can be computed using the Pythagorean theorem and adding the sides of the rectangle.

Rectangle:
Width = 4 - 2 = 2 meters 
Length = 11 - 2 = 9 meters

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long leg of the triangle = length of the rectangle = 9 meters.

12² + 9² = c²
144 + 81 = c² 
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15 = c

Point A to B = 4 - 2 = 2 meters
Point B to C = 15 meters
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Point D to A = 11 - 2 = 9 meters

Total distance traveled = 2 + 15 + 14 + 9 = 40 meters.
6 0
3 years ago
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).

5 0
3 years ago
Help 35 points, this is algebra 1 on system of equations and inequalities
Rama09 [41]

Answer:

idk how to do that try to slove it

but I didn't get right answer

4 0
3 years ago
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