Answer:
A
Step-by-step explanation:
To determine if the ordered pairs lie on the graph, substitute the x- coordinate of the point into the equation and if the value agrees with the y- coordinate of the point then it lies on the graph
A
x = - 2 : y = -2(- 2) + 7 = 4 + 7 = 11 ≠ 3 ← (-2,3) not on graph
B
x = - 1 : y = - 2(- 1) + 7 = 2 + 7 = 9 ← (- 1, 9) lies on graph
C
x = 3 : y = -2(3) + 7 = - 6 + 7 = 1 ← (3, 1) lies on graph
D
x = 4 : y = - 2(4) + 7 = - 8 + 7 = - 1 ← (4, - 1) lies on graph
I hope you want to simplify this expression.
To simplify the given expression we can use the order of operation.
So, first step is to operate multiplication. Therefore,
400 x 34 = 13600
Hence,
400x34+23444444444-23444444444
= 13600 +23444444444-23444444444
= 13600 + 0 Since +23444444444-23444444444 = 0
= 13600.
So, the simplified value of the given expression is 13600.
Hope this helps you!.
Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.
X = 112
See attachment file below.
Hope it helped!
