P'Q' is 4 cm
A'B' is 3 cm
M'N' is 6 cm
After the scale factor is applied to the line segments .
Hope I helped!
Add three to both sides
2x greater than 14 - 5x
Add 5x to both sides
7x greater than 14
divide both sides by 7
x is greater than 2
The answer to your question is y=-3
<h2>
Answer:</h2>
Tessellation is possible only for equilateral triangles, squares and regular hexagons. Therefore, there are only three regular polygons that tessellate.
Tessellation is possible, if each interior angle of a polygon is a factor of 360 degrees.
Now, in the given image, pentagon and octagon cannot tessellate.
And the triangle does not seem to be an equilateral triangle. But in general we can say that triangle tessellates.
So, the answer might be triangle (if its equilateral)
The required equation of the hyperbola is expressed as
![\frac{x^2}{10}- \frac{y^2}{15}=1 \\](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%7D%7B10%7D-%20%5Cfrac%7By%5E2%7D%7B15%7D%3D1%20%5C%5C)
The standard form for calculating the equation of a parabola along the x-axis is expressed as:
where:
(h, k) is the center
(k±c, h) are the foci
is the directrix
From the question given, we can see that foci = (±5, 0)
k±c, h = ±5, 0
k = 0
h = 0
c = 5
From the directrix,
![x=h\pm\frac{a^2}{c}\\x =0\pm\frac{a^2}{5}\\\pm2 = \pm\frac{a^2}{5}\\a^2 = 10\\](https://tex.z-dn.net/?f=x%3Dh%5Cpm%5Cfrac%7Ba%5E2%7D%7Bc%7D%5C%5Cx%20%3D0%5Cpm%5Cfrac%7Ba%5E2%7D%7B5%7D%5C%5C%5Cpm2%20%3D%20%5Cpm%5Cfrac%7Ba%5E2%7D%7B5%7D%5C%5Ca%5E2%20%3D%2010%5C%5C)
Also, we need to know that;
a²+b² = c²
10 + b² = 5²
b² = 25 - 10
b² = 15
Substituting the gotten values into the equation of a hyperbola;
![\frac{(x-0)^2}{10}- \frac{(y-0)^2}{15} =1\\\frac{x^2}{10}- \frac{y^2}{15} =1\\](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-0%29%5E2%7D%7B10%7D-%20%5Cfrac%7B%28y-0%29%5E2%7D%7B15%7D%20%3D1%5C%5C%5Cfrac%7Bx%5E2%7D%7B10%7D-%20%5Cfrac%7By%5E2%7D%7B15%7D%20%3D1%5C%5C)
This gives the required equation of the hyperbola
Learn more on the equation of hyperbola here: brainly.com/question/20409089