Answer: the answer is 3
Step-by-step explanation:
An increase of 100% in the value of (rh) means the value doubles. When that doubled value is squared, the new area is 4 times the old area.The question asks how many times GREATER the new A is than the old A. 4 times AS LARGE AS is a 300% INCREASE, which is 3 TIMES LARGER THAN.So the grammatically correct answer is 3
Answer:
Step-by-step explanation:
6
Answer:
Option b is correct (8,13).
Step-by-step explanation:
7x - 4y = 4
10x - 6y =2
it can be represented in matrix form as![\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-4%5C%5C10%26-6%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
A= ![\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-4%5C%5C10%26-6%5Cend%7Barray%7D%5Cright%5D%20)
X= ![\left[\begin{array}{c}x\\y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D)
B= ![\left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
i.e, AX=B
or X= A⁻¹ B
A⁻¹ = 1/|A| * Adj A
determinant of A = |A|= (7*-6) - (-4*10)
= (-42)-(-40)
= (-42) + 40 = -2
so, |A| = -2
Adj A= ![\left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-6%264%5C%5C-10%267%5Cend%7Barray%7D%5Cright%5D%20)
A⁻¹ = / -2
A⁻¹ = ![\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-2%5C%5C5%26-7%2F2%5Cend%7Barray%7D%5Cright%5D%20)
X= A⁻¹ B
X= ![\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-2%5C%5C5%26-7%2F2%5Cend%7Barray%7D%5Cright%5D%20%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%283%2A4%29%20%2B%20%28-2%2A2%29%5C%5C%285%2A4%29%20%2B%20%28-7%2F2%2A2%29%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}12-4\\20-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12-4%5C%5C20-7%5Cend%7Barray%7D%5Cright%5D)
X= ![\left[\begin{array}{c}8\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
x= 8, y= 13
solution set= (8,13).
Option b is correct.
We know that
case 1)
Applying the law of sines
a/Sin A=b/Sin B
A=56°
a=12
b=14
so
a*Sin B=b*Sin A----> Sin B=b*Sin A/a---> Sin B=14*Sin 56°/12
Sin B=0.9672
B=arc sin (0.9672)------> B=75.29°-----> B=75.3°
find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+75.3)----> C=48.7°
find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 48.7°/Sin 56°)
c=10.87-----> c=10.9
the answer Part 1)
the dimensions of the triangle N 1
are
a=12 A=56°
b=14 B=75.3°
c=10.9 C=48.7°
case 2)
A=56°
a=12
b=14
B=180-75.3----> B=104.7°
find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+104.7)----> C=19.3°
find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 19.3°/Sin 56°)
c=4.78-----> c=4.8
the answer Part 2)
the dimensions of the triangle N 2
are
a=12 A=56°
b=14 B=104.7°
c=4.8 C=19.3°
