All categories

3 years ago

Find the quotient. Round to the nearest tenth. 3784 divided by 18

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

5 0

Answer:

210.22

Step-by-step explanation:

first you need to divide 3785 divided by 18 which is 210.22222222 then u need to round to the nearest tenth which is equaled to 210.22. I hope this helped

You might be interested in

You scored a 91 on a test with bonus points. you got 4 bonus points how many points did you score before the bonus points

Alex Ar [27]

You need to subtract the bonus points form the final score: 91-4=87

3 0

3 years ago

Read 2 more answers

Please tell me if my answer is right <br> NO LIES brainlest<br> report you then

ladessa [460]

Answer:

Your answers are correct.

Step-by-step explanation:

The "<" sign means less than and the ">" sign means greater than. It is clear that you have a handle on them.

1.) If Steve has no more that 28 toys, then the less than sign would represent the fact that Steve can have a maximum of 28 toys, and no more.

2.) If the temperature is more that 28 degrees Fahrenheit, then the greater than sign is appropriate in order to imply that the variable represents a higher temperature.

3.) If tony is younger than 29 years old, then the less than sign fits to show that the variable represents an age that is smaller than 29.

4.) If the table is greater than 29 kilograms, then the greater than sign is needed. It shows that the table has a greater weight than 29 kilograms.

Hope this helps! :D

4 0

3 years ago

A cylinder has a 12-inch diameter and is 15 inches tall. It is filled to the top with water. A 6-inch-diameter ball is placed wi

Tems11 [23]

The volume of the cylinder is the space occupied by the cylinder. The volume of the water in the cylinder is 1583.36266 in³.

<h3>What is the volume of a cylinder?</h3>

The volume of the cylinder is the space occupied by the cylinder. It is calculated with the help of the formula,

$\text{Volume of the cyllinder}= \pi r^2h$

As it is given that the diameter of the cylinder is 12 inches while the height of the cylinder is 15 inches, also, it has a ball of the diameter of 6 inches placed inside the cylinder, therefore, the volume of the water that is inside the cylinder is the difference in the volume of the cylinder and the volume of the spherical ball.

$\text{Volume of the cyllinder}= \pi r^2h = \dfrac{\pi}{4}d^2 h$

Substitute the values,

$\begin{aligned}\text{Volume of the cyllinder}&= \dfrac{\pi}{4}d^2 h\\& = \dfrac{\pi}{4} \times (12^2) \times 15\\ &=1696.46\rm\ in^3\end{aligned}$

Now, the volume of the ball is equal to the volume of the sphere therefore, the volume of the ball can be written as,

$\text{Volume of the sphere} = \dfrac{4}{3}\pi r^3$

$= \dfrac{4}{3}\pi r^3\\\\= \dfrac{4}{3}\times \pi \times (3^3)\\\\= 113.097\rm\ in^3$

Further, the volume of the water that is inside the cylinder can be written as,

The volume of water = Volume of the cylinder - Volume of the sphere

= 1696.46 - 113.097

= 1583.36266 in³

Hence, the volume of the water in the cylinder is 1583.36266 in³.

Learn more about Volume of the Cylinder:

brainly.com/question/1780981

5 0

2 years ago

1. Andrew ran a marathon in about 1.56 x 10^4 s. (a) Write Andrews time in standard notation. (b) Evelyn ran a marathon in 4 h a

romanna [79]

Answer:

(a) 15,600 s
(b) Evelyn

Step-by-step explanation:

(a) 1.56×10^4 = 1.56×10000 = 15,600

(b) There are 3600 seconds in an hour, so Andrew's time was

15600 s/(3600 s/h) = 13/3 h = 4 1/3 h

1/3 hour is 20 minutes, so Evelyn's time of 4:16 is shorter than Andrew's time of 4:20.

7 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago