Round 72.807 to the nearest tenth.

72.81
72.9
70
72.8

Answers

Answer 1
Answer:

Answer:

72.8

Step-by-step explanation:

The tenth is the immidiate number after the decimal point, so if you are rounding tho the nearest tenth, you have to remember that if you have any number from 5-9 you have to round up, and if you have any number from 1-4 you have to round down, in this case as you have a .807, it´s a ,07 you have to round down to 72.8

Answer 2
Answer:

Answer:

answer 72.8

Step-by-step explanation:


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Can someone help me solve 5(t+2) please

Answers

This is a simple multiplication, written in a strange form.
 5(t+2) means 5 x t + 5 x 2 which simplifies to:
5t + 10
If it's equal to a number, 0 for example, then with a little rearranging we can find what t is.
5t + 10 = 0,
5t = -10
t = -2
If it's not equal to something, the answer is just 5t + 10.

What does 5^(3*5^(-4)) equal plus steps

Answers

If you simplify the expression, you get 5 3/625.
In decimal form, your answer would be 1.00775521...
5^3(5^−4)

=5^3(1/625)
=5^3/625

=1.007755

or the function f given above, determine whether the following conditions are true. Input T if the condition is ture, otherwise input F . (a) f′(x)<0 if 00 if x>2; F (c) f″(x)<0 if 0≤x<1; T (d) f″(x)>0 if 14; T (f) Two inflection points of f(x) are, the smaller one is x= 2 and the other is x= 2

Answers

In linear equation, the two inflection points of f(x) are at    x = 1    or   x = 4

What are instances of linear equations?

  • Ax+By=C is the typical form for linear equations involving two variables.
  • A standard form linear equation is, for instance, 2x+3y=5. It is rather simple to locate both intercepts when an equation is stated in this way (x and y).
  • When resolving systems of two linear equations, this form is also incredibly helpful.

We are aware that f(x) has a decreasing order if f'(x) 0 and a rising order if f'(x) > 0.

seen in the graph If f(x) is evidently growing on x (2, ) and decreasing on x ( 0,2) Since f(x) is dropping, if 0 x 2 is true, then f'(x) 0.

Since f(x) is rising, f'(x) > 0 if x > 2 is true.

Since f(X) is concave down, if 0 x 1 is true, then f"(x) 0 and vice versa.

Since f(X) is concave up, f"(x) > 0 and is true if 1 x 4

Given that f(X) has a concave downward shape, e) f"(x) 0 if x > 4 is true.

The two inflection points of f(x) are at x = 1 or x = 4.

Learn more about linear equation here:

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Answer:

0 if 0≤x<1; T (d) f″(x)>0 if 14; T

Step-by-step explanation:

2 and the other is x= 2

A square has an area of 9 cm2 what is its side length​

Answers

Value of side length is 3 cm

Give that;

Area of square = 9 cm²

FInd:

Value of side length

Computation:

We know that,

Area of square = Side²

BY putting value into formula

⇒ Side² = 9

⇒ Side² = 3²

Side = 3

So,

Value of side length = 3 cm

Learn more about square;

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Answer:

Length side is 3 cm

Step-by-step explanation:

the length in square is called side.

A= sides X sides

9 = s^2

S= 3

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HELP!!!!!!!! The histogram below shows the estimated monthly salaries of company employees with different years of experience. How might the graph be misleading?

Answers

The scale on the x-axis is not completed for the expected monthly salary.

What is a histogram?

A diagram is made up of rectangles whose size is related to a variable's frequency and whose length is similar to the measurement range.

The histogram below shows the estimated monthly salaries of company employees with different years of experience.

The scale for the graph's x-axis (i.e. Years of Experience) does not completely account for the expected monthly salary (scale on the y-axis).

More about the histogram link is given below.

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Answer:  
_________________________________________________________
The "scale" used on the "x-axis" for the graph shown (i.e. "Years of Experience") does not fully account for the "estimated monthly salaries" (scale on the "y-axis")  AMONG different experience levels .
_________________________________________________________

Ex 2.8
3. find the maximum value of the curve y=x²-4x+4 for -3≤x≤3

Answers

y=x^2-4x+4\ny'=2x-4\n\n2x-4=0\n2x=4\nx=2\n2\in[-3,3]\n\ny''=2>0 \Rightarrow \text{ There's a minimum at } x=2

In this case you have find the values of y(-3) and y(3).

y(-3)=(-3)^2-4\cdot(-3)+4=9+12+4=25\ny(3)=3^2-4\cdot3+4=9-12+4=1\n\ny(-3)>y(3) \Rightarrow y_(max)=25