Let the angle be x and it's complement be c.
Then, x = c + 88 and x + c = 90
Substitute the first equation in the second.
(c+88) + c = 90
2c+88 = 90
2c = 2
c =1
Compliment = 1
Angle = 89
Answer:

Step-by-step explanation:
We do not have enough information for slope intercept form. But we can use the point-slope formula to find the information. The formula is
where we substitute a point (x,y) for
.
We have m=3/4 and (4, 1). We input m and
.

We now simplify the parenthesis and solve for y.

We convert -4 into a fraction with 1 as the denominator.

We add 1 to both sides to isolate y,

This is slope intercept form. The line as slope 3/4 and y-intercept (0,2) or b=2.
Answer:
D is not a function.
Step-by-step explanation:
The first option is a function because it makes up a straight line. The second option is also a function because it is also known as y = -3/2x + 2 creating a slope function. The third option is also a function because there is no more than one point for each x value. The last option is not a function because the number 2 is going for 2 variables A and B and you can't have them with more than one so D is your answer.
Answer:
d
Step-by-step explanation:
in ''d)'', all the numbers are irrational
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=
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=
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.