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2 years ago

9

Rob's tablet is fully charged.He uses 1/12 of the charge playing games,5/12 of the charge reading and 3/12 completing homework.W

hat fraction of the charge remains on Rob's tablet

1 answer:

ozzi2 years ago

3 0

All of the number already have the same denominator, which is 12 so you can just add them.

(1+5+3)/12=9/12

and subtract then from 1 = 12/12
(12-9)/12=3/12

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Use the substitution method to solve the following system of equations: 3x + 5y = 3 x + 2y = 0 (1 point) (1, 1) (6, −3) (6, 1) (

PtichkaEL [24]

Step-by-step explanation:

Given.

$3x+5y=3$

$x+2y=0$

Take the second equation and subtract 2y to both sides.

$(x+2y)-2y=0-2y$

$x=-2y$

Substitute x into the first equation and simplify.

$3x+5y=3$

$3(-2y)+5y=3$

$-6y+5y=3$

$-y=3$

Invert.

$y=-3$

Substitute Y into your second equation.

$x+2y=0$

$x+2(-3)=0$

$x-6=0$

Add -6 to both sides.

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$x=6$

Answer:

(6, -3)

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lapo4ka [179]

Answer:

He is a soccer player

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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)

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

$(x,y)$ → $(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

$R=(2,1)$

$R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

$R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})$

$R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})$

$R'=(2,3)$

$S=(2,-2)$

$S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$S'=(2,-2)$

$T=(-1,-2)$

$T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

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

Now we can finally draw our triangle:

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