Identify the coordinates of point A after a translation with rule

Mathematics

2 answers:

den301095 [7]3 years ago

8 0

To identify the new coordinates after the translation I need the old coordinates

Bond [772]3 years ago

3 0

Your answers are simple they are (3,2), (2,3), (1,-2), and (1,2). I hope the answers help. ☺

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the product of two consecutive positive even integers is 14 more than their sum. Set up an equation that can

alexandr1967 [171]

Required equation to determine two consecutive even number is $x^{2}-16=0$ and required numbers are 4 and 6.

<u>Solution:</u>

Given that product of two consecutive positive integer is 14 more that their sum. Need to create the equation and solve it to get two numbers.

Let’s assume first even number be represented by x.

So second consecutive even number will be $x + 2$

Sum of two consecutive number = $(x) + (x + 2)$

$\Rightarrow$ Sum of two consecutive number = $2x + 2$

Product of two consecutive number = $(x) \times(x+2)=x^{2}+2 x$

Product of two consecutive number is 14 more that sum of two consecutive numbers, so if we subtract 14 from product, we will get sum.

$\begin{array}{l}{\Rightarrow\left(x^{2}+2 x\right)-14=2 x+2} \\\\ {=>x^{2}+2 x-2 x-14-2=0} \\\\ {\Rightarrow x^{2}-16=0} \\\\ {\Rightarrow x^{2}-4^{2}=0} \\\\ {\Rightarrow x^{2}=4^{2}} \\\\ {\Rightarrow x=4}\end{array}$

One even number = $x=4$

Other even number = $x + 2 = 4 + 2 = 6$

Hence required equation to determine two consecutive even number is $x^2 -16 = 0$ and required numbers are 4 and 6.

8 0

4 years ago

Customers per day: 98, 87, 93, 82, 101, 99, 97, 97, 91<br> Find the mean to the nearest tenth:

notsponge [240]

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7 0

3 years ago

Read 2 more answers

EFGH is a rhombus. Given EG = 16 and FH = 12, what is the length of one side of the rhombus?

Black_prince [1.1K]

Answer:

Step-by-step explanation:

It is given that EFGH is a rhombus and EG = 16 and FH = 12.

We know that the diagonals of the rhombus are the perpendicular bisectors, therefore OF=6, OH=6, OE=8 and OG=8.

Now, using the Pythagoras theorem in ΔOFG, we have

$(FG)^{2}=(OG)^{2}+(OF)^{2}$