What is the Greatest Common Factor of 48r^2-52r ?

What is the midpoint of 18.6 and 18.5?

Nina [5.8K]

The midpoint of 18.6 and 18.5 is 18.55

5 0

3 years ago

Please help I’m really need the help (NO LINKS ) (NO LYING )

MAVERICK [17]

Answer:

for this you should use photomath all you do is take a picture of each problem and put the answer down

7 0

3 years ago

Read 2 more answers

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:

8 0

3 years ago

Whats the answer i need help

julia-pushkina [17]

Answer)3.5
1)minus 17.5-14=3.5
2)14+3.5=17.5
3)keep adding 3.5

6 0

3 years ago

|1/2b-8|=|1/4b-1|<br> b=____ and ____

Deffense [45]

Answer:

b = 12 and 28

Step-by-step explanation:

The absolute value equation |1/2b-8| = |1/4b-1| resolves to a piecewise linear function with three pieces. There are two solutions.

<h3>Domains</h3>

The absolute value function on the left has a turning point where its value is zero:

1/2b -8 = 0

b -16 = 0

b = 16

The absolute value function on the right has a turning point where its value is zero:

1/4b -1 = 0

b -4 = 0

b = 4

For b > 16, both absolute value functions are identity functions. In this domain, the equation is ...

1/2b -8 = 1/4b -1

For 4 < b < 16, the function on the left negates its argument, so the equation in this domain is ...

-(1/2b -8) = 1/4b -1

For b < 4, both functions negate their arguments, so the equation in this domain is ...

-(1/2b -8) = -(1/4b -1)

For the purpose of finding the value of b, this is effectively identical to the equation for b > 16. (The value of b does not change if we multiply both sides of the equation by -1.)

<h3>Solutions</h3>

<u>Domain b < 4 ∪ 16 < b</u>

1/2b -8 = 1/4b -1

2b -32 = b -4 . . . . . . . . multiply by 4

b = 28 . . . . . . . . . . . . add 32-b to both sides

This solution is in the domain of the equation, so is one of the solutions to the original equation.

<u>Domain 4 < b < 16</u>

-(1/2b -8) = 1/4b -1 . . . . equation in this domain

-2b +32 = b -4 . . . . . . multiply by 4

36 = 3b . . . . . . . . . . . add 2b+4 to both sides

12 = b . . . . . . . . . . . . divide by 3

This solution is in the domain of the equation, so is the other solution to the original equation.

<h3>Graph</h3>

For the purposes of the graph, we have define the function g(b) to be the difference of the two absolute value functions. The solutions are found where g(x) = 0, the x-intercepts. The graph shows those to be ...

b = 12 and b = 28

__

<em>Additional comment</em>

Defining g(b) = |1/2b-8| -|1/4b-1|, we can rewrite it as ...

$g(b)=\begin{cases}7-\dfrac{1}{4}b&\text{for }b < 4\\-\dfrac{3}{4}b+9&\text{for }4\le b < 16\\\dfrac{1}{4}b-7&\text{for }16\le b\end{cases}$

Then the solutions are the values of b that make g(b) = 0.

4 0

2 years ago