Answer:
![m=-\frac53; q=-2](https://tex.z-dn.net/?f=m%3D-%5Cfrac53%3B%20q%3D-2)
Step-by-step explanation:
Let's bring the line in the classical
form
![3y= -5x-6\\y=-\frac53 x -2](https://tex.z-dn.net/?f=3y%3D%20-5x-6%5C%5Cy%3D-%5Cfrac53%20x%20-2)
At this point we have the two values
for the slope and
for the intercept
|x-6| < 1 + 10 (move 10 to the right side)
|x-6| < 11
1)x-6 < 11 (here x >= 6)
2)-x+6 < 11 (here x < 6)
1)x<17
2)x>-5
so we get -5 < x < 17
the answer: (-5; 17)
1. n^2 -8n +16 = 25
Subtract 25 from both sides
n^2 - 8n + 16 - 25 = 0
Simplify
n^2 - 8n - 9 =0
Factor
(n-9)(n+1) = 0
Solve for n
n-9 = 0, n = 9
n+1 = 0, n = -1
Solution: 9,-1
2. C = b^2/25
Multiply both sides by 25:
25c = b^2
Take square root of both sides
b = +/-√25c
Simplify:
b = 5√C, -5√C
3. d = 16t^2 +12t
subtract d from both side:
16t^2 + 12t -d =0
Use quadratic formula to solve:
t = (3 +/-√(9-4d))/8
4. 5w^2 +10w =40
Subtract 40 from both side:
5w^2 + 10w -40 = 0
Factor:
5(w-2)(w+4)=0
Divide both sides by 5:
(w-2)(w+4)=0
Solve for w:
w-2 = 0, w = 2
w+4=0, w = -4
Solution: 2,-4
Answer:
9cm
Explicación:
El teorema de pitagoras nos dice que la hipotenusa al cuadrado es igual a la suma de los catetos al cuadrado:
![c^{2} =a^{2} + b^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D%20%3Da%5E%7B2%7D%20%2B%20b%5E%7B2%7D)
Donde c es la hipotenusa y a y b son los catetos.
Por lo cual, podemos reemplazar c por 15cm y a por 12cm :
![15^{2} =12^{2} +b^{2}](https://tex.z-dn.net/?f=15%5E%7B2%7D%20%3D12%5E%7B2%7D%20%2Bb%5E%7B2%7D)
Finalmente , debemos resolver la ecuación para b, así que b es igual a :
![225=144+b^{2} \\225-144=144+b^{2} -144\\81=b^{2} \\\sqrt{81} =b\\9=b](https://tex.z-dn.net/?f=225%3D144%2Bb%5E%7B2%7D%20%5C%5C225-144%3D144%2Bb%5E%7B2%7D%20-144%5C%5C81%3Db%5E%7B2%7D%20%5C%5C%5Csqrt%7B81%7D%20%3Db%5C%5C9%3Db)
Por lo tanto, la longitud del otro cateto es 9 cm
Answer:
-11/45
Step-by-step explanation:
5/9 + (-4/5)
25/45 + (-36/45)
(Multiplied each side by opposite denom.)
25/45 -36/45
Simplify
25-36=-11
-11/45
Cannot simpfily