Answer:
Step-by-step explanation:
(4 , -3) ; (8 , -12)
![Slope =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\dfrac{-12-[-3]}{8-4}\\\\=\dfrac{-12+3}{4}\\\\=\dfrac{-9}{4}](https://tex.z-dn.net/?f=Slope%20%3D%5Cdfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-12-%5B-3%5D%7D%7B8-4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-12%2B3%7D%7B4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-9%7D%7B4%7D)
m = -9/4 ; (4 , -3)
![y - y_{1}= m(x -x_{1}\\\\y - [-3]= \dfrac{-9}{4}(x -4)\\\\\\y + 3 = \dfrac{-9}{4}x -4*\dfrac{-9}{4}\\\\\\y + 3 =\dfrac{-9}{4}x+9\\\\\\y = \dfrac{-9}{4}x+9-3\\\\y=\dfrac{-9}{4}x+6](https://tex.z-dn.net/?f=y%20-%20y_%7B1%7D%3D%20m%28x%20-x_%7B1%7D%5C%5C%5C%5Cy%20-%20%5B-3%5D%3D%20%5Cdfrac%7B-9%7D%7B4%7D%28x%20-4%29%5C%5C%5C%5C%5C%5Cy%20%2B%203%20%3D%20%5Cdfrac%7B-9%7D%7B4%7Dx%20-4%2A%5Cdfrac%7B-9%7D%7B4%7D%5C%5C%5C%5C%5C%5Cy%20%2B%203%20%3D%5Cdfrac%7B-9%7D%7B4%7Dx%2B9%5C%5C%5C%5C%5C%5Cy%20%3D%20%5Cdfrac%7B-9%7D%7B4%7Dx%2B9-3%5C%5C%5C%5Cy%3D%5Cdfrac%7B-9%7D%7B4%7Dx%2B6)
Answer:
The answer is the first option.
Step-by-step explanation:
When you flip it over the y axis. A would still be the same. ANd the first one matches A.
Answer:
The length of the sides of the right angle triangle
15 , 25, 20
Step-by-step explanation:
<u>Explanation</u>:-
<u>Step(i):-</u>
we know that by using Pythagorean theorem to find the lengths of the sides of the triangle
AC² = AB² + BC²
Given a right triangle has legs labeled 3m and 2m+10
let us assume that AB = 3m and BC = 2m + 10
Given a hypotenuse labeled 5m
let us assume that hypotenuse AC = 5m
<u>Step(ii)</u>:-
Now by using Pythagorean theorem
AC² = AB² + BC²
(5m)² = (3m)² + (2m+10)²
25m² = 9m² + 4m² + 40m + (10)² ( since (a + b)² = a²+2ab+b²) )
on simplification , we get
25m²-13m² -40m -100 =0
12m² -40m -100 =0
4(3m² -10m -25) =0
3m² -10m -25 =0
3m² - 15m + 5m -25 =0
3m(m-5) + 5(m-5) =0
(3m +5) (m-5) =0
3m +5 =0 and m-5=0
3m = -5 and m =5
and m=5
we can not choose negative value
so the value m=5
<u>Step (iii)</u>:-
The sides of right angle triangle
AB = 3m
AB = 3(5) = 15 and
BC = 2m + 10
BC = 2(5) +10 = 20
The hypotenuse AC = 5m
AC = 25
<u>Conclusion:</u>-
The lengths of the sides of the right triangle
15, 25 ,20
(1) <em>f(x)</em> = <em>x</em>² - 4<em>x</em> + 2, <em>g(x)</em> = 3<em>x</em> - 7
<em>(f</em> o<em> g) (x)</em> = <em>f(g(x))</em> = <em>f</em> (3<em>x</em> - 7)
= (3<em>x</em> - 7)² - 4(3<em>x</em> - 7) + 2
= (9<em>x</em>² - 42<em>x</em> + 49) + (-12<em>x</em> + 28) + 2
= 9<em>x</em>² - 57<em>x</em> + 79
(2) <em>g(x)</em> = -6<em>x</em> + 5, <em>h(x)</em> = -9<em>x</em> - 11
<em>(g</em> o <em>h) (x)</em> = <em>g(h(x))</em> = <em>g</em> (-9<em>x</em> - 11)
= -6(-9<em>x</em> - 11) + 5
= 54<em>x</em> + 66 + 5
= 54<em>x</em> + 71
(3) <em>f(x)</em> = √(2<em>x</em> - 5), <em>g(x)</em> = 5<em>x</em>² - 3
<em>(g</em> o <em>f) (x)</em> = <em>g(f(x))</em> = <em>g</em>(√(2<em>x</em> - 5))
= 5 (√(2<em>x</em> - 5))² - 3
= 5 (2<em>x</em> - 5) - 3
= 10<em>x</em> - 25 - 3
= 10<em>x</em> - 28