Write the log equation as an exponential equation. You do not need to solve for x.

PLEASE FOLLOW THE DIRECTIONS AND SHOW YOUR WORK. IM OFFERING 15PTS (Way more than its worth) AND BRAINLIEST ANSWER. PLEASE HELP

Alex

I'll take this challenge since I'm not feeling lazy for once.... x would represent Gavin and y would represent Seiji.

a. $x+y=425, x-y=25$

b. I will solve your system using substitution.
$x+y-425, x-y=25$
Step 1: Solve $x+y=425$ for x
$x+y+-y=425+-y$ (add -y for both sides)
$x=-y+425$
Step 2: substitute $-y+425$ for x in $x+b=25$
$x+y=25$
 $-y+425-y=25$
$-2y+425=25$ (simplify both sides of the equation)
$-2y+425+-425=25+-425$ (add -425 to both sides)
 $-2y=-400$
$\frac{-2y}{-2} = \frac{-400}{-2}$
y=200
step 3: substitute 200 for y in $x=-y+425$
$x=-y+425$
$x=-200+425
$
$x=225$
$x=225, y= 200$

c. Already answered with the picture that is attached.

d. Gavin earned 225 dollars while Seiji earned 200 dollars.

3 0

3 years ago

Quadratic Formula to solve the equation x^2- 4x = - 7

mylen [45]

roots of the equation x^2- 4x = - 7 are 2+3i and 2-3i

What are the natures of root of quadratic equation?

Case I: $4ac > b^2$

The roots of the quadratic equation $ax^2 + bx + c = 0$ are real and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive.

Situation II: $b^2$ $-4ac$ =0

The roots and of the quadratic equation $ax^2 + bx + c = 0$ are real and equal when a, b, and c are real numbers, a 0, and the discriminant is zero.

Case III: $b^2$ $-4ac$ <0

The quadratic equation $ax^2 + bx + c = 0$ has roots when a, b, and c are real numbers, a 0, and the discriminant is negative, but these roots are not equal and are not real. We refer to the roots in this instance as fictitious.

Case IV: Perfect square and $b^2 - 4ac > 0$

The roots of the quadratic equation $ax^2 + bx + c = 0$ are real, rational, and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive and perfect square.

Quadratic Formula $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$