Graph and solve y+3X=5

Answers

Answer 1

Answer: $y+3x=5\n\ny=-3x+5\n\nx=0\ \ \ \Rightarrow\ \ \ y=-3\cdot 0+5=5\n\ny=0\ \ \ \Rightarrow\ \ \ 0=-3x+5\ \ \ \Rightarrow\ \ \ 3x=5\ \ \ \Rightarrow\ \ \ x= (5)/(3)$

$the\ graph\ in\ annex$

Related Questions

The digit in the tens place is twice as big as the digit in the tens place in 182,437. A) True B) False
Solve the following equations. 1 − log8(x − 3) = log8(2x)
a boat traveled 336 miles downstream and back.the trip downstream took 12 hours.the trip back took 14 hours.what is the speed of the boat in still water?what is the speed of the.current?
Which of the following is the reciprocal of x2 – 2x – 8?
Evaluate j/4 when j=12

Convert to vertex form.
y=2x^2+14x-4

Answers

$y=2x^2+14x-4\n\na=2;\ b=14;\ c=-4\n\nvertex\ form:y=a(x-h)^2+k\n\nwhere:h=(-b)/(2a)\ and\ k=(-(b^2-4ac))/(4a)\n\nh=-(-14)/(2\cdot2)=-(7)/(2)\n\nk=(-(14^2-4\cdot2\cdot(-4)))/(4\cdot2)=(-(196+32))/(8)=(-228)/(8)=-(57)/(2)\n\n\nAnswer:y=2(x+(7)/(2))^2-(57)/(2)$

$To \ convert \ the \ standard \ form \ y = ax^2 + bx + c \ of \ a \ function \ into \ vertex \n \nform \ y = a(x - h)^2 + k \n \n Here \ the \ point \ (h, k) \ is \ called \ as \ vertex \n \n h=(-b)/(2a) , \ \ \ \ k= c - (b^2)/(4a)$

$y=2x^2+14x-4 \n \na=2 ,\ b=14 , \ c=-4 \n \n h=(-14)/(2*2)=-(14)/(4)=-3.5 \n \nk= -4 - (14^2)/(4\cdot 2)=-4-(196)/(8)=-4-24.5=-28.5 \n \n y=2(x+3.5)^2 -28.5$

Find the common ratio of the sequence: -75, -15, -3, -0.6, . . .

Thank you!

Answers

5:1 is the ratio. You get this by dividing 75/15, 15/3, 3/0.6...

Compare and contrast the absolute value of a real number to that of a complex number.

Answers

The absolute value of a real number is a positive value of the number. Which means that the absolute value is the distance from zero of the number line. However, that of the complex numbers is the distance from the origin to the point in a complex plane.

The definition of complex, real and pure imaginary number is as follows:

$A \ \mathbf{complex \ number} \ is \ written \ in \ \mathbf{standard \ form} \ as:\n \n \ (a+bi) \n \n where \ a \ and \ b \ are \ real \ numbers. \ If \ b=0, \ the \ number \ a+bi=a \n is \ a \ \mathbf{real \ number}. \ If \ b\neq 0, \ the \ number \ (a+bi) \ is \ called \ an \n \mathbf{imaginary \ number}. \ A \ number \ of \ the \ form \ bi, \ where \ b\neq 0, \n is \ called \ a \ \mathbf{pure \ imaginary \ number}$

The absolute value of this number is given by:

$|a+bi|=\sqrt{a^(2)+b^(2)}$

So, the absolute value of a complex number represents the distance between the origin and the point in the complex plane. On the other hand, the absolute value of a real number means only how far a number is from zero without considering any direction.

Pls help me my homework is due tomorrow lol