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lawyer [7]
3 years ago
7

What is the diameter of a circle that has an area of 452.16 mm 2

Mathematics
2 answers:
Papessa [141]3 years ago
7 0
C = pi * d
452.16 = 3.14 * d
Divide both sides by 3.14
144 = d
kipiarov [429]3 years ago
5 0

stop with the wrong answer's it is

24mm

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Please Help Me With This problem
evablogger [386]

Answer:

250

Step-by-step explanation:

First, we need to find h°g, which is the same thing as h(g(x))

This means that we will take the function h(x) and will replace each instance of x with x^2+x-6 as that is the value of g(x)

h(x)=5x\\\\h(g(x))=5g(x)\\\\h(g(x))=5(x^2+x-6)\\\\h(g(x))=5x^2+5x-30

Now that we have our composed function, we can find the y-value when x=7

h(g(7))=5(7^2)+5(7)-30\\\\h(g(7))=5(49)+35-30\\\\h(g(7))=245+5\\\\h(g(7))=250

5 0
3 years ago
Suppose y = f (x) +k what effect does k have on parent funcation
Aliun [14]
K is the amount that the function will go up or down.
It is typically the y intercept
8 0
3 years ago
Select all expressions that represent the area of the given triangle
Mice21 [21]

Answer:

A=1/2(12x16)

Step-by-step explanation:

because it's the area of the square divided by two

3 0
3 years ago
For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant
nirvana33 [79]

Answer:

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of y with respect to x over the interval by definition of secant line:

r = \frac{y(b) -y(a)}{b-a} (1)

Where:

a, b - Lower and upper bounds of the interval.

y(a), y(b) - Function exaluated at lower and upper bounds of the interval.

If we know that y = 3\cdot x^{2}, a = 3 and b = 6, then the average rate of change of y with respect to x over the interval is:

r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}

r = 27

The average rate of change of y with respect to x over the interval [3,6] is 27.

b) The instantaneous rate of change can be determined by the following definition:

y' =  \lim_{h \to 0}\frac{y(x+h)-y(x)}{h} (2)

Where:

h - Change rate.

y(x), y(x+h) - Function evaluated at x and x+h.

If we know that x = 3 and y = 3\cdot x^{2}, then the instantaneous rate of change of y with respect to x is:

y' =  \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}

y' =  3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}

y' = 3\cdot  \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}

y' = 6\cdot  \lim_{h \to 0} x +3\cdot  \lim_{h \to 0} h

y' = 6\cdot x

y' = 6\cdot (3)

y' = 18

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

5 0
3 years ago
Which of the following show an element of the sample space for first rolling a die and then tossing a coin?
Sphinxa [80]
A because if you use T3 it wouldn’t have make sense so only reasonable and solution that works is A
5 0
2 years ago
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