f the polygon is translated 4 units down and 5 units right, what will the coordinates of the new image be

Mathematics

1 answer:

Diano4ka-milaya [45]4 years ago

5 0

I know how to figure out the answer though I am at school right now, I will be home in about 20 minutes. I will answer first thing when I get back.

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Write an inequality that would include the point (3,78), (0,312), (5,10) to be part of the solution.

just olya [345]

Answer:

We want to write an inequality with the points (3, 78), (0, 312) and (5, 10) as solutions.

Then the set of the x-values is {0, 3, 5}

The set of the y-values is {10. 78. 312}

So if we wrote an inequality like:

y > 5

It would include all these solutions, because:

10 > 5

78 > 5

312 > 5

And this has no restriction on the x-variable, so all the x-value range can be a solution, meaning that the points (3, 78), (0, 312) and (5, 10) are solutions to that inequality.

Another one (also trivial) can be:

x < 9

because:

3 < 9

0 < 9

5 < 9

And we have no restriction on the y-values, then the points (3, 78), (0, 312) and (5, 10) are solutions to that inequality.

8 0

3 years ago

PLZ HELP ASAP

svet-max [94.6K]

Answer is D:
e + 5 = 3e - 11

2e = 16
e = 8

Ella is 8 years old
Jamie is 8+5 = 13 years old

6 0

3 years ago

5x when x=7 ......... NEED HELP

Gnoma [55]

Answer:

Step-by-step explanation:

5 x 7 = 35

5 0

4 years ago

Read 2 more answers

Calculate this reflection of the triangle:

Mice21 [21]

Answer:

a = 3, b = 0, c = 0, d = -2

Step-by-step explanation:

<em>To find the reflection Multiply the matrices</em>

∵ The dimension of the first matrix is 2 × 2

∵ The dimension of the second matrix is 2 × 3

<em>1. Multiply the first row of the 1st matrix by each column in the second matrix add the products of each column to get the first row in the 3rd matrix.</em>

2. Multiply the second row of the 1st matrix by each column in the second matrix add the products of each column to get the second row of the 3rd matrix

$\left[\begin{array}{ccc}1&0\\0&-1\end{array}\right]$ × $\left[\begin{array}{ccc}0&3&0\\0&0&2\end{array}\right]$ = $\left[\begin{array}{ccc}(1*0+0*0)&(1*3+0*0)&(1*0+0*2)\\(0*0+-1*0)&(0*3+-1*0)&(0*0+-1*2)\end{array}\right]=\left[\begin{array}{ccc}0&3&0\\0&0&-2\end{array}\right]$