Answer:
Three squares that form right triangle:
A, E and C
Step-by-step explanation:
Area of E = 1^2 = 1
Area of A = 2.4^2 = 5.76
Area of C = 2.6^2 = 6.76
Area A + Area E = Area C
Solve by elimination ie eliminate x or y from the equations by performing operations on them.
first label the equations , to follow the process.
x - y = 5 ----------------(1)
x+ y = 3 ----------------(2)
If (1) and (2) are added then y will be eliminated.
(1) + (2) gives : 2x = 8 → x = 4
now substitute this value of x into either of the 2 equations and solve for y.
let x = 4 in (1) : 4 - y = 5 → -y = 1 → y = -1
check in (1) : 4-(-1) = 4+1 = 5
check in(2) : 4 - 1 = 3
So your answer is x = 4 , y = -1
Answer:
x = 49
Step-by-step explanation:
Let's solve for x in the first triangle by applying the leg rule.
Leg rule is given as: hypotenuse/leg = leg/part
Hypotenuse = x
Leg = 28
Part = 16
Plug in the values into the equation
x/28 = 28/16
Cross multiply
x*16 = 28*28
16x = 784
x = 784/16
x = 49
✔️Let's check our answer to see if we would get the same value of x for the other triangle. Apply the same leg rule.
Hypotenuse = x
Leg = 21
Part = 9
Thus:
x/21 = 21/9
Cross multiply
x*9 = 21*21
9x = 441
x = 441/9
x = 49
✔️We have the same value. Therefore, x = 49 is the correct answer.
Step-by-step explanation:
if I understand this correctly than you are looking for the inverse function of f(x) = y = 3x³.
the inverse function simply tries to calculate the original x it of the original y.
and then, to make it a formal function, we rename x to y and y to x.
y = 3x³
x³ = y/3
![x = \sqrt[3]{ \frac{y}{3} }](https://tex.z-dn.net/?f=x%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7By%7D%7B3%7D%20%7D%20)
=> as "regular" function
![y = {f}^{ - 1} (x)= \sqrt[3]{ \frac{x}{3} }](https://tex.z-dn.net/?f=y%20%20%3D%20%20%7Bf%7D%5E%7B%20-%201%7D%20%28x%29%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7Bx%7D%7B3%7D%20%7D%20)
Step-by-step explanation:
<h3 /><h3>D)X=-4, -2</h3>
is the answer