| u | = √(2² + (-1²)) = √5
| v | = √ ( 1² + (-8)² = √65
cos (u,v) = ( u * v ) / (| u | * | v |) =
(2 * 1 + ( -1 ) * ( - 8 )) / √5 √ 65 = (2 + 8) / √5 √65 = 10 / (√5 √ 65 )
The length of a larger diagonal:
d 1² = | u |² + 2 |u| |v| + | v |² = 5 + (2 √5 √65 * 10 / √5 √65 )+65
d 1² = 70 + 20 = 90
d 1 = √ 90 = 3√10
d 2² = 70 - 20 = 50
d 2 = √50 = 5√2
Answer:
The lengths of the diagonals are: 3√10 and 5√2 .
Answer:
45 hours on math, 20 on science, 18 on soc st, and 17 on la
Step-by-step explanation:
percents are how many out of one hundred
The answer is 110 because if you multiply 8•16=128 and 18 cars are used, you subtract 128-18 which is 110.
<span>First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.
You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...
The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span>

<span> where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.</span>