Answer:
False
Step-by-step explanation:
it depends on the person, if you try to boast your confidence it you may either be more confident or you will lose all of your current confidenc.
Answer:
3rd choice 0.57
Step-by-step explanation:
when rounding to the tenth place we look at the hundreds place and when it greater than or at 5 we round up and when it's less than 5 we round down.
1) explanation:0.7 when rounding to the tenth place. since the 8 was greater than 5 we round-up and get 0.70.
2)explanation: 0.5 when rounded to the tenth place the hundreds place which is 2 is less the 5 so we round down and get 0.50.
3)explanation: 0.6 when rounded its hundreds place is greater than 5 so we round up and get 0.60.
4)explanation: 0.7 when rounding to the tenth place. since the 9 was greater than 5 we round-up and get 0.70.
hopes this help you
Money Required for making 1 Sandwich that is Cost Price of 1 Sandwich = S Dollar
Selling Price of 1 Sandwich = 20 % more than Cost Price S
= S + 20 % of S
= 
=$ 1.20 S
= $(S + 0.20 S)
Selling Price of 73 Sandwiches = $ 73 (S + 0.20 S)
The expression $73(S+0.2 S) →→ Selling Price of 73 Sandwiches
Differentiate g using the power rule:




The critical points of g occur where g' is zero or undefined.
We have

and the derivative is undefined for

We will need to shade the region above f(x) and the region below the function g(x).
<h3>How to transform the graph into the solution set?</h3>
We have:
f(x) = x² - 2
g(x) = -x² + 5
Both of these are already graphed, and we want to transform it into:
y > f(x)
y ≤ g(x)
The first inequality means that we need to graph f(x) with a dashed line, because f(x) is not part of the solution, and then we shade all the region above f(x).
For the other inequality, we use a solid line (because the points on the line are solutions) and then we shade the part below the curve.
Learn more about inequalities on:
brainly.com/question/18881247
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