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1 year ago

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Which phrase best describes f(x), graphed on the coordinate plane below?64(0)2-х2-46Of(x) is an odd function.Of(x) is an even fu

nction.Of(x) is both odd and even.Of(x) is neither odd nor even.

1 answer:

Marianna [84]1 year ago

8 0

Functions which satisfy the following:

$f(-x)=-f(x)$

are odd functions

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Please help me will mark as brainlist

Nookie1986 [14]

Answer:

17.3825

Step-by-step explanation:

First I figured out the area of the square which is 18×18=324

Then I found out the are of the circle which is 254.47, so 324 - 254.47 = 69.53

Then there are 4 times the shaded are so divided by 4, so 69.53 ÷ 4 = 17.3825

PLEASE GIVE BRAINLIEST!

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4 years ago

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NEED HELP I SUCK AT GEOMETRY (OFFERING 100POINTS)!!! best answer get brainlest

Olegator [25]

Answer: This is kinda confusing but I would say no

Step-by-step explanation: Because they would be congruent if the little line in JKD is in the middle of JK instead of KD.

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3 years ago

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7th grade math help me pleasee

elena55 [62]

Answer:

No

Step-by-step explanation:

I don't think that's how those specifc diagrams work. X in this case would be 3 not seven.

Yes you can move the x and the 1s but it would break the whole diagram. It would just be an (expression/term). Not a full diagram.

You said this was 7th grade math? Wow. I learned this in 5th grade!

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3 years ago

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Simplify 4^-4.<br> please help I have 10 minutes

kolezko [41]

Answer:

1/256

Step-by-step explanation:

$4^{-4}$

$(1/4)^{4}$

=> 1/256

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3 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

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3 years ago

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