5x + 3y = - 53
the equation of a line in ' slope- intercept form ' is y = mx + c
where m is the slope and c the y-intercept
rearrange 3x - 5y = - 15 into this form to obtain m → (subtract 3x from both sides)
- 5y = - 3x - 15 → divide all terms by - 5 )
y =
x + 3 → in slope-intercept form with m = 
given a line with slope m then the slope m₁ of a line perpendicular to it is
m₁ = -
= - 1 ÷
= - 
partial equation is y = -
x + c
to find c substitute ( - 7, - 6) into the partial equation
- 6 =
+ c ⇒ c = - 6 -
= - 
y = -
x -
→ in slope intercept form
multiply all terms by 3
3y = - 5x - 53 → ( add 5x to both sides )
5x + 3y = - 53 → in standard form
The answer is 13. To solve this you have to do 4*3 which is 12. Then you subtract 25 and 12 to get an answer of 13.
Answer:
take 40 degree as reference angle
using sine rule
let opposite be x
sine 40=opposite/hypotenuse
0.64=x/7
0.64*7=x
4.48=x
Step-by-step explanation:
Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
