I need to understand proofs

using the midpoint formula, solve for the midpoint of the segment with these pairs of endpoints (-11,30) and (-26,-2)

REY [17]

Answer:

Midpoint is $(\frac{-37 }{2}, 14)$

Step-by-step explanation:

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2})$

= $(\frac{-11 -26}{2}, \frac{30 - 2 }{2})$

= $(\frac{-37 }{2}, \frac{28 }{2})$

= $(\frac{-37 }{2}, 14)$

Midpoint is (-37/2, 14)

3 0

2 years ago

PLZ SOMEONE HELP ME AND EXPLAIN <br> THX

ELEN [110]

The graph of g is one-fifth as steep as the graph of f.

The function g basically takes the inputs for f and multiplies them by one-fifth, which means the outputs are one-fifth times those of f. Multiplying by one-fifth makes something smaller (it's the same as dividing by five). It helps to visualize this relationship, so I've attache the graphs below.

7 0

2 years ago

Find the side length of a. Pls help!

sdas [7]

Answer:

im too eionf0qn3490fn0wiopf

Step-by-step explanation:

6 0

1 year ago

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

2 years ago

What's the missing Vertical angles

skelet666 [1.2K]

1.) <1=25°, <2=155°, <4=155°

5 0

2 years ago

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