'One third' as a decimal is 1 ÷ 3 = 0.333333..... ⇒ The decimal is recurring
'One fifth' as decimal is 1 ÷ 5 = 0.2 ⇒ The decimal terminates at the tenth value
'One seventh' as decimal 1 ÷ 7 = 0.142857142857.... ⇒ The decimal is recurring for every digit between 1 and 7
'One ninth' as decimal 1 ÷ 9 = 0.111111.... ⇒ The decimal is recurring
Answer: one fifth
900 - 250 = 650
650 / 250 = 13 / 3 = 2.6 = 260%
Answer:
h(x) = 2^(x) - 1.
Step-by-step explanation:
Let's look at each equation:
f(x) = -3x +7, Well as x increases, since it's multiplication, there are "going to be more" -3's, so it's going to be decreasing.
g(x) = -4(2^x). While 2^x is increasing, because "there are going to be more 2's multiplied by each other" as x increases, it's being multiplied by a negative number, so it's actually going to be decrasing
h(x) = 2^(x) - 1. Here's it's going to be increases as x goes towards infinity because "there are going to be more 2's multiplied by each other", and there isn't any negative sign, while there is a negative 1, it's constant, so the overall value will be increasing
Answer:
a) 3 b) 5 c) 7 d) 9
Step-by-step explanation:
For this, you want to replace x in the equation y=x+5 with each of the values listed.
For the first one, -2, the equation becomes y=-2+5, which is solved to y=3.
For the second one, 0, the equation becomes y=0+5, which is solved to y=5.
For the third one, 2, the equation becomes y=2+5, which is solved to y=7.
For the last one, 4, the equation becomes y=4+5, which is solved to y=9.
**This content involves solving algebraic equations with a known variable, which you may wish to revise. I'm always happy to help!
