Answer:
Step-by-step explanation:
two points on the line are (-3,-2) and (5,4)
slope=(4+2)/(5+3)=6/8=3/4
eq. of line is
y+2=3/4 (x+3)
4y+8=3x+9
4y=3x+1
y=3/4 x+1/4
Answer: x = 8
(the correct answer was not provided as an option)
<u>Step-by-step explanation:</u>
![log_2(x-6)+log_2(x-4)=log_2(x)\\\\log_2[(x-6)(x-4)]=log_2(x)\\\\(x-6)(x-4)=x\\\\x^2-10x+24=x\\\\x^2-11x+24=0\\\\(x-3)(x-8)=0\\\\x=3\qquad x=8\\\\\\\text{The term after the log symbol (inside the parenthesis) must be greater than 0!}\\\\Check:\\3-6>0\ \text{FALSE --- so 3 is not a valid solution}\\\\8-6>0\ \text{TRUE}\\8-4>0\ \text{TRUE}\\8>0\ \text{TRUE --- 8 is a valid solution because it is true for all}](https://tex.z-dn.net/?f=log_2%28x-6%29%2Blog_2%28x-4%29%3Dlog_2%28x%29%5C%5C%5C%5Clog_2%5B%28x-6%29%28x-4%29%5D%3Dlog_2%28x%29%5C%5C%5C%5C%28x-6%29%28x-4%29%3Dx%5C%5C%5C%5Cx%5E2-10x%2B24%3Dx%5C%5C%5C%5Cx%5E2-11x%2B24%3D0%5C%5C%5C%5C%28x-3%29%28x-8%29%3D0%5C%5C%5C%5Cx%3D3%5Cqquad%20x%3D8%5C%5C%5C%5C%5C%5C%5Ctext%7BThe%20term%20after%20the%20log%20symbol%20%20%28inside%20the%20parenthesis%29%20must%20be%20greater%20than%200%21%7D%5C%5C%5C%5CCheck%3A%5C%5C3-6%3E0%5C%20%5Ctext%7BFALSE%20---%20so%203%20is%20not%20a%20valid%20solution%7D%5C%5C%5C%5C8-6%3E0%5C%20%5Ctext%7BTRUE%7D%5C%5C8-4%3E0%5C%20%5Ctext%7BTRUE%7D%5C%5C8%3E0%5C%20%5Ctext%7BTRUE%20---%208%20is%20a%20valid%20solution%20because%20it%20is%20true%20for%20all%7D)
The new temperature is 60 degrees
The volume of a box with dimensions x by x by x is

.
So if we are given the volume V of a cube-shaped box, the side length of thit is
![\sqrt[3]{V}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7BV%7D%20)
.
So we calculate the cubic roots of the volumes we have, and we add their heights.
To calculate the cubic root of 729, we can factorize it, and group the perfect cubes together, as follows:


, which we recognize as

so

similarly 1,331 can be found to be

.
Thus we have 2 boxes with side length equal to 11 m and one with side length equal to 9 m.
Answer: h= 11+11+9 = 31 (meters)
Answer:
<em>It has been given that Rectangle Q has an area of 2 square units.</em>
<em>Thea Drew a scaled version of Rectangle Q and marked it as R.</em>
<em>As you must keep in mind If we draw scaled copy of pre-image, then the two images i.e Pre-image and Image are similar.</em>
<em>As you have not written what is the scale factor of transformation</em>
<em>Suppose , Let the Scale factor of transformation= k</em>
<em>Rectangle Q = Pre -image, Rectangle R= Image</em>
<em>If, Pre-Image < Image , then scale factor is k >1.</em>
<em>But If, Pre-Image > Image , then Scale factor will be i.e lies between, 0<k<1.</em>