Write a real world word problem that contains y=3x+4 and y=2x+6

Mathematics

1 answer:

hjlf3 years ago

7 0

Answer:

Step-by-step explanation:

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Find the domain of the function.

SpyIntel [72]

Domain means the values that can be used as a function's input. Since all real values of x will yield a real value of f(x), the domain is (-infinity, infinity).

6 0

2 years ago

Consider the following system of equations:

IrinaK [193]

Answer:

x = 6 ; y = -9

Step-by-step explanation:

8 x + 2 y = 30 ..........equ No 1

7 x + 2 y = 24......... equ No 2

8 x = 30 - 2y

∴ x = $\frac{ 30 - 2 y}{8}$

substituting the value of x in equ No 2

7 $\frac{30 - 2 y}{8}$ + 2 y = 24

7 ( 30 -2 y) + 2 y × 8 = 24 × 8

7 × 30 - 14 y + 16 y = 192

210 + 2 y = 192

2 y = 192 - 210

2 y = - 18

∴ y = - 9

put y = -9 , 8 x = 30 - 2 y

8 x = 30 - 2 ( -9)

8 x = 30 - ( -18)

8 x = 48

∴ x = 6

x = 6 ; y = -9

3 0

3 years ago

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ

Deffense [45]

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$