Answer:
43cm²
Step-by-step explanation:
let's first consider the area of a square.
the area is L² which means all sides are equal so we take the length times the breadth which is both equal because like we said all sides are equal.
so to find the side of the square using the area, we take the square root of both of the area.
and also
so we have the height of the triangle as 5cm and the base is 4.2cm.
now, from the triangle, since we have two sides and it's a right-angled, we can use Pythagoras' formula.
so the side 6.53cm is also the same side as the largest triangle. Since it's a square, all sides are equal. So we find the area of the largest triangle by using the formula
Area = L²
Area = 6.53²
Area = 42.6cm
the nearest cm square
Area = 43cm²
Answer:
-4/5
Step-by-step explanation:
r=x/2
a=2x
ar²=-1/16
x^3=-1/8
x=-1/2
sn=a/1-r
=-4/5
To solve the problem, substitute the given points for x in the given equation to get
Solving the three equations simultaneously, we have:
a = -3, b = 2 and c = -5
Therefore, the required equation is
Answer:
<h3>★ <u>11/30</u> is the right answer. ★</h3>
Step-by-step explanation:
- Number of male students who got 'A' in the test is 11
- Number of female students who got 'A' in the test is 19
- Total students who got 'A' in the test is 30
- Probability that the male student got an 'A' is P(A | male) = (Number of male students who got 'A' in the test)/(Number of total students who got 'A' in the test) = <em><u>11/30</u></em>
Answer:
Option C is correct.
Step-by-step explanation:
-x+3y=2
4x-2y=22
In matrix form is represented as:
![\left[\begin{array}{cc}-1&3\\4&-2\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{c}2&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%263%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2622%5Cend%7Barray%7D%5Cright%5D)
AX=B
|A| = (-1)(-2)-(3)(4)
|A| = 2-12
|A| = -10
Adj A = ![\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-3%5C%5C-4%26-1%5Cend%7Barray%7D%5Cright%5D)
A^-1 = -1/10
A^-1 = 1/10![\left[\begin{array}{cc}2&3\\4&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%263%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D)
X= A^-1 B
X = 1/10![\left[\begin{array}{c}2&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2622%5Cend%7Barray%7D%5Cright%5D)
X=1/10![X=1/10\left[\begin{array}{c}2*2+3*22\\4*2+1*22\end{array}\right]\\X=1/10\left[\begin{array}{c}4+66\\8+22\end{array}\right]\\X=1/10\left[\begin{array}{c}70\\30\end{array}\right]\\X=\left[\begin{array}{c}70/10\\30/10\end{array}\right]\\X=\left[\begin{array}{c}7\\3\end{array}\right]](https://tex.z-dn.net/?f=X%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2A2%2B3%2A22%5C%5C4%2A2%2B1%2A22%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%2B66%5C%5C8%2B22%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D70%5C%5C30%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D70%2F10%5C%5C30%2F10%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D7%5C%5C3%5Cend%7Barray%7D%5Cright%5D)
So, x = 7 and y =3
Hence Option C is correct.
