Two less than three-fifths a number less or equal to forty-one

Mathematics

2 answers:

Ivan2 years ago

8 0

Answer:

$\frac{3}{5}x-2\leq 41$

$x\leq \frac{215}{3}$

Step-by-step explanation:

In inequality,

'≥' = greater than equal to,

'≤' = less than equal to,

Given phrase,

two less than three-fifths a number less or equal to forty-one,

Let x represents the unknown number,

two less than three-fifths of a x = $\frac{3}{5}x-2$

$\frac{3}{5}x-2$ less or equal to forty-one,

$\frac{3}{5}x-2\leq 41$

Which is the required inequality,

By solving it,

$\frac{3}{5}x\leq 43$

$x\leq \frac{43\times 5}{3}=\frac{215}{3}$

Leviafan [203]2 years ago

6 0

(x divided by 3/5) -2<41
-
i hope this helps

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Please help me <br> Show your work <br> 10 points

Svet_ta [14]

<h2>Answer</h2>

After the dilation $\frac{5}{3}$ around the center of dilation (2, -2), our triangle will have coordinates:

$R'=(2,3)$

$S'=(2,-2)$

$T'=(-3,-2)$

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

$(x,y)$ → $(x-2, y+2)$

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor $\frac{5}{3}$ . Therefore our second partial rule will be:

$(x,y)$ → $\frac{5}{3} (x-2,y+2)$

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )$

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)$