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Zolol [24]
3 years ago
12

Peter walks 30 feet away from his house and places a mirror on the ground. He backs 6 feet away from the mirror so that he can s

ee the tip of the roof. Peter's eyes are 5 feet above the ground. Peter and the house are both perpendicular to the ground. The angles between the top of the house, the mirror, and the ground and between Peter's eyes, the mirror, and the ground are congruent as shown in the image below:
What is the height of the house? Show your work and explain your reasoning in complete sentences. PLEASE HELP THANKS

Mathematics
1 answer:
Natali [406]3 years ago
4 0
Work is in the attached picture. Whenever you are given a problem like this, draw it out. It makes it a lot easier!

The house is 25 feet tall.

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24. q + r t s, for q = 46, r = 19. s = 54​
Kaylis [27]

(24 \times 46 )+ (19 \times t \times 54) =

1104 + 1026t

_________________________________

And we're done.

Thanks for watching buddy good luck.

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8 0
3 years ago
A vector is in standard position, with its terminal point in the second quadrant and an x-coordinate of -5. The vector
Nonamiya [84]

Answer:6.0, 130

Step-by-step explanation:

Just got the question right on edgenuity.

3 0
3 years ago
Log(x-2)+log2=2logy<br>log(x-3y+3)=0​
In-s [12.5K]

Answer:

<h2>x = 4 and y = 2 or x = 10 and y = 4</h2>

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
3 years ago
Mega Movies hosted a film premiere on Friday night. They charged $9 for adults and $6 for children. One hundred twenty-seven adu
Oxana [17]

x = number of adult tickets

y = number of children tickets

9x + 6y = 1053

x+y = 127 so x = 127 - y

replace x = 127 - y into 9x + 6y = 1053

9(127 - y) + 6y = 1053

1143 -9y + 6y = 1053

-3y = 1053 -1143

-3y = -90

y = 30

x = 127 - 30

x = 97

answer: 97 adults and 30 children




3 0
3 years ago
At the store Jason finds the total volume of soda in a pack is 50.24 cubic inches . If each cylinder -shaped can has a height of
atroni [7]

Answer: A) 4

Step-by-step explanation:

Hi, to answer this, first, we have to calculate the volume of each can:

Volume of a cylinder: π x radius^2 x height

Since:

Diameter: 2 radius

2 = 2r

2/2 =r

r=1

Back with the volume formula:

V = 3.14 ( 1)² 4 =12.54 cubic inches

Now, we have to divide the total volume of soda by the volume of each can:

50.24/12.54= 4 soda cans

Feel free to ask for more if needed or if you did not understand something.

7 0
3 years ago
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