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3 years ago

WILL MARK BRAINLIEST!!!

1 answer:

4 0

$f(x) = -9x^4 + 5x + 3\\\\ f(-x)= -9(-x)^4 + 5\cdot(-x) + 3=-9x^4-5x+3\\\\ f(-x)\not =f(x)\implies \text{ not even}\\\\ -f(x)=-( -9x^4 + 5x + 3)= 9x^4 - 5x - 3\\\\ -f(x)\not =f(-x) \implies \text{ not odd}$

Neither

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What is the value of the expression 5/7 divided by 8/9

kirill115 [55]

45/56
5/7 divided 8/9
5/7 times 9/8
45/56

7 0

3 years ago

In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each

Rudiy27

Answer:

$N(t) = 972(1.015)^{t}$

Growth function.

The number of students enrolled in 2014 is 1162.

Step-by-step explanation:

The number of students in the school in t years after 2002 can be modeled by the following function:

$N(t) = N(0)(1+r)^{t}$

In which N(0) is the number of students in 2002 and r is the rate of change.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

In 2002, there were 972 students enrolled at Oakview High School.

This means that $N(0) = 972$

Since then, the number of students has increased by 1.5% each year.

Increase, so r is positive. This means that $r = 0.015$

Then

$N(t) = N(0)(1+r)^{t}$

$N(t) = 972(1+0.015)^{t}$

$N(t) = 972(1.015)^{t}$

Growth function.

Find the number of students enrolled in 2014.

2014 is 2014-2002 = 12 years after 2002, so this is N(12).

$N(t) = 972(1.015)^{t}$

$N(12) = 972(1.015)^{12}$

$N(12) = 1162$

The number of students enrolled in 2014 is 1162.

7 0

3 years ago

Does someone know the answer?

Drupady [299]

Answer:

i do not

know the awnser but c would be 720 multiplied by its self 3 times the converted

Step-by-step explanation:

8 0

3 years ago

Nina can ride her bike 63,360 feet in 3,400 seconds, and Sophia can ride her bike 10 miles in 1 hour. What is Nina's rate in mil

Sloan [31]

Nina's rate is 12.7 mph and Sophia's rate is 10 mph. Therefore, Nina is faster.

5 0

3 years ago

Read 2 more answers

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago