Ask question

Ask question

All categories

3 years ago

5

Which of the following statements is not true?

1 answer:

8_murik_8 [283]3 years ago

7 0

In order for us to know which statements are true, we need to go over the choices.

A. True
0.6 = 3/5
0.6 = 3/5
60 / 100 = 3 / 5

B. True
0.125 = 1/8
0.125 = 125/1000 = 1/8

C. FALSE.
2.015 = 2 15/1000 = 2 3/200

The statement that is false is Choice <span>C. 2.015 = 2 1/100</span>

You might be interested in

Emma jogs 2 miles along the beach in

Juli2301 [7.4K]

Answer:

<h3><u>3 times the time --> 3 times the distance</u></h3>

Step-by-step explanation:

Emma jogs 2 miles along the beach in 1/3 of an hour. If she travels at a constant rate, how far will she jog in an hour?

6 0

2 years ago

Read 2 more answers

Lady bird [3.3K]

Small triangle : (1,4), (3,1), (3,4)
The x: (3,2)
The big triangle: (-3,3), (-1,5), (-2,1)

(For the big triangle that is only correct if the black dot is the mirror line)

4 0

3 years ago

Read 2 more answers

What is 180 divided by 20,835

grigory [225]

$\frac{180}{20,835} = \frac{4}{463} = 0.00863930885$

3 0

3 years ago

Read 2 more answers

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

b.

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

c.

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

6 0

3 years ago

Read 2 more answers

Dr.Purple wants to buy candy bars for all his 143 math students. He buys 9 cases plus 8 more individual bars. How many bars are

Annette [7]

There is 15 bars in each case.

Subtract 8 from 143 and you get 135.

Divide 135 by 9 and you get 15 :)

3 0

2 years ago

Read 2 more answers