Answer:

Step-by-step explanation:
b = 4 × 3.5 = 14 cm²
h = 2,1 cm
then



Answer:
They the answernis 17.50 fam
Step-by-step explanation:
We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:
<h3>
Presenting the equation:</h3>
Basically, we want to solve:
![\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%262%5C%5Ca%261%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Db%26c%5C%5C1%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix product will be:
![\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-b%20%2B%202%26-c%20%2B%202d%5C%5Ca%2Ab%20%2B%201%26a%2Ac%20%2B%20d%5Cend%7Barray%7D%5Cright%5D)
Then we must have:
-b + 2 = 1
This means that:
b = 2 - 1 = 1
We also need to have:
a*b + 1 = 0
we know the value of b, so we just have:
a*1 + b = 0
Now the two remaining equations are:
-c + 2d = 0
a*c + d = 1
Replacing the value of a we get:
-c + 2d = 0
-c + d = 1
Isolating c in the first equation we get:
c = 2d
Replacing that in the other equation we get:
-(2d) + d = 1
-d = 1
Then:
c = 2d = 2*(-1) = -2
So the values are:
If you want to learn more about systems of equations, you can read:
brainly.com/question/13729904
X-1/x= 2
x-1=2x
x-2x=1
-x=1
x=-1
x+1/x
-1+1/-1
0/-1
0
it equals 0
It's a parallelogram so if one angle is 90 degrees they all will be because of transverse angles and all that good stuff.
So we're given the diagonal of a rectangle and one side and we're asked to find the other. The diagonal of a rectangle is the hypotenuse of the right triangle whose legs are the sides of the rectangle. So this is a Pythagorean Theorem question in disguise:




Answer: 140 cm
I don't recall seeing this Pythagorean Triple before.