Answer:
Yes they will intersect
Function 1= F(X)=2X+5
Function 2=H(X)=3X+2
INTERSECT=(3,11)
Step-by-step explanation:
First of all, we create 2 LINEAR function, i created the function f(x)=2x+5 and the function h(x)=3x+2, both are linear(without a quadratic term). Then
you replace the x for a number:
Table 1 (F(X)=2X+5) Table 2 (H(X)=3X+2)
X=1----->Y=2+5=7 X=1------>Y=3·1+2=5
X=2---->Y=2·2+5=9 X=2----->Y=3·2+2=8
X=3---->Y=3·3+5=11 X=3----->Y=3·3+2=11
With both tables of data we can see that in the X=3/Y=11 point this two linear functions will intersect so the answer is that the two functions will intersect at (3,11)----->(X,Y)
Answer:
x = 4
y = 1
Step-by-step explanation:
x = 3y + 1 (Multiply everything by -2)
2x + 4y = 12 (Transfer 4y to the right of the = sign)
2x = -4y + 12
-2x = -6y - 2
0 = -10y + 10
-10 = -10y
y = 1
x = 3y + 1
x = 3(1) + 1
x = 3 + 1
x = 4
![\bf \cfrac{x}{4x+x^2}\implies \cfrac{\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}{\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(4+x)}\implies \cfrac{1}{4+x}\qquad \{x|x\in \mathbb{R}, x\ne -4\}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7Bx%7D%7B4x%2Bx%5E2%7D%5Cimplies%20%5Ccfrac%7B%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D%7D%7B%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%284%2Bx%29%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B4%2Bx%7D%5Cqquad%20%5C%7Bx%7Cx%5Cin%20%5Cmathbb%7BR%7D%2C%20x%5Cne%20-4%5C%7D)
if you're wondering about the restriction of x ≠ -4, is due to that would make the fraction with a denominator of 0 and thus undefined.
Answer :
To find the product of (x + 8)(x – 10)
We know that
(x + a)(x + b) = x2 + (a + b)x + ab
In the expression (x + 8)(x – 10)
a = 8
b = -10
a + b = 8- 10 = -2
ab = 8 × (-10) = – 80
Substitute the values in the above equation
(x + 8)(x – 10) = x2 + (8 – 10)x + 8 × (-10)
(x + 8)(x – 10) = x2 – 2x – 80.
Thank you !
Answer: -36a^6+32a^4
Step-by-step explanation:
