Answer:
12,000
Step-by-step explanation:
i believe this would be what she withdrew but if not i'm sorry for giving the wrong answer
An expo fn, such as y = e^x, has neither max nor min. The graph of y = e^x begins in Quadrant II and ends in Quadrant I; it has no upper limit. As x becomes smaller and smaller, the graph approaches, but does not touch, the x-axis.
FALSE
Answer:
x=−4 and y=1
Step-by-step explanation:
Rewrite equations:
y=−2x−7;−3x+4y=16
Step: Solvey=−2x−7for y:
y=−2x−7
Step: Substitute−2x−7foryin−3x+4y=16:
−3x+4y=16
−3x+4(−2x−7)=16
−11x−28=16(Simplify both sides of the equation)
−11x−28+28=16+28(Add 28 to both sides)
−11x=44
-11x/-11=44/-11
x=−4
Step: Substitute−4forxiny=−2x−7:
y=−2x−7
y=(−2)(−4)−7
y=1(Simplify both sides of the equation)
x=−4 and y=1
Answer:
3
Step-by-step explanation:
8 ÷ 2² + 1
8 ÷ 4 + 1
2 + 1
3
Have a good day!
Step-by-step explanation:
3=j
so 36(j) now divide that 36 divided 3 = 12
so 12=j
(This was kind of confusing but this is the best I can give you!)
