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Marta_Voda [28]
3 years ago
12

Find the area of a circle of radius 4 cm.​

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
4 0

Answer:

Step-by-step explanation:

Area of a circle=πr²

A=π*4²

A=16π

A=50.27 ( rounded to nearest 100th)

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Tomas has $1,000 to spend on a vacation. His
Jobisdone [24]

Answer:

He can spend $118.50 each day

Step-by-step explanation:

1000 - 348.25 = 651.75

651.75 / 5.5 = 118.5

4 0
3 years ago
To obtain an average (arithmetic mean) of exactly 2/3, what fraction must be added to 1/6, 1/2, and 1/3?
ollegr [7]

Answer:

D. 5/3

Step-by-step explanation:

\frac{1/6+1/2+1/3+x}{4} =2/3

1/6+1/2+1/3+x=8/3

x=8/3-1/6-1/2-1/3

x=8/3-1/6-3/6-2/6

x=8/3-6/6

x=16/6-6/6=10/6

simplified

x=5/3

Hope this helps

3 0
1 year ago
What is the best measurement to eastimate the volume of a juice box
denis-greek [22]

1 : you would use a graduated cylinder . 2 : , you would use a yard stick or maybe a ruler if you are not able to find one. You can also use a meter stick.

7 0
2 years ago
Please help, need for math hw and cant figure it out
swat32

The standard form of the quadratic function f(x) = -3x^2 + 6x - 2 is f(x) = -3x^2 + 6x - 2

<h3>How to represent the quadratic function in standard form?</h3>

The quadratic function is given as

f(x) = -3x^2 + 6x - 2

The standard form of a quadratic function is represented as:

f(x) = ax^2 + bx + c

When both equations are compared, we can see that the function f(x) = -3x^2 + 6x - 2 is already in standard form

Where

a = -3

b = 6

c = -2

Hence, the standard form of the quadratic function f(x) = -3x^2 + 6x - 2 is f(x) = -3x^2 + 6x - 2

Read more about quadratic function at

brainly.com/question/25841119

#SPJ1

7 0
1 year ago
For what value of c is the function defined below continuous on (-\infty,\infty)?
kozerog [31]
f(x)= \left \{ {{x^2-c^2,x \ \textless \  4} \atop {cx+20},x \geq 4} \right&#10;

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if 
</span><span>\lim_{x \rightarrow 4} \  f(x) = f(4)

</span><span>In notation we write respectively
</span>\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence 
\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2

Thus these two limits, the one from above and below are equal if and only if
 4c + 20 = 16 - c²<span> 
 Or in other words, the limit as x --> 4 of f(x) exists if and only if
 4c + 20 = 16 - c</span>²

c^2+4c+4=0&#10;\\(c+2)^2=0&#10;\\c=-2

That is to say, if c = -2, f(x) is continuous at x = 4. 

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers (-\infty, +\infty)

4 0
3 years ago
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