All categories

3 years ago

What is the relationship between the 6s in the number 7,664

Mathematics

1 answer:

Xelga [282]3 years ago

4 0

The answer to your question is 600 and 60

You might be interested in

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B60%7D%20%7D%7B%5Csqrt%5B3%5D%7B36%7D%20%7D" id="TexFormula1" title=

Margarita [4]

∛60
--------
∛36

=
∛5
------
∛3

=∛45
-----------
3

= 1.185

3 0

3 years ago

Read 2 more answers

Sue has $1.15 in nickels and dimes, a total of 16 coins. How many nickels and how many dimes does Sue have?

adell [148]

0.05x+0.10y=1.15
x+y=16
solve the system

6 0

3 years ago

Read 2 more answers

What is the area of this shape?

zysi [14]

Area = L x H

area = 20 x 15 = 300 square feet

3 0

3 years ago

There are 30 puzzle books on a shelf. The number of puzzle books is 25 of all the books on the shelf.

Softa [21]

The total number of books on the shelf is 12 books

Given the following parameters

Total number of puzzle books on the shelf = 30 books

If the number of puzzle books is 25 of all the books on the shelf, the total number of books on the shelf is expressed as:

= 2/5 of 30

= 2/5 * 30

= 2 * 6

= 12

Hence the total number of books on the shelf is 12 books

Learn more here: brainly.com/question/13885389

6 0

3 years ago

Read 2 more answers

If dy dx equals cosine squared of the quantity pi times y over 4 and y = 1 when x = 0, then find the value of x when y = 3.

Afina-wow [57]

Answer: $x = -\frac{8}{\pi}$

Explanation: Note that

$\frac{dy}{dx} = \cos^2 \left ( \frac{\pi y}{4} \right )
\\
\\ \frac{dy}{\cos^2 \left ( \frac{\pi y}{4} \right )} = dx
\\
\\ \int{\frac{dy}{\cos^2 \left ( \frac{\pi y}{4} \right )}} = \int dx
\\
\\ \boxed{x = \int{\sec^2 \left ( \frac{\pi y}{4} \right )dy}}$

To evaluate the integral in the boxed equation, let $u = \frac{\pi y}{4}$ . Then,

$du = \frac{\pi}{4} dy 
\\ \Rightarrow \boxed{dy = \frac{4}{\pi}du}$

So,

$x = \int{\sec^2 \left ( \frac{\pi y}{4} \right )dy}
\\
\\ = \int{\sec^2 u \left ( \frac{4}{\pi}du \right )}
\\
\\ = \frac{4}{\pi}\int{\sec^2 u du}
\\
\\ = \frac{4}{\pi} \tan u + C
\\
\\ \boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) + C}\text{ (1)}$

Since y = 1 when x =0, equation (1) becomes

$0 = \frac{4}{\pi} \tan \left ( \frac{\pi (1)}{4} \right ) + C 
\\ 
\\ 0 = \frac{4}{\pi} (1) + C 
\\
\\ \frac{4}{\pi} + C = 0
\\
\\ \boxed{C = -\frac{4}{\pi}}$

With the value of C, equation (1) becomes

$\boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} }$

Hence, if y = 3,

$x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} 
\\
\\ x = \frac{4}{\pi} \tan \left ( \frac{\pi (3)}{4} \right ) -\frac{4}{\pi} 
\\
\\ x = \frac{4}{\pi} (-1) -\frac{4}{\pi} 
\\
\\ \boxed{x = -\frac{8}{\pi}}$

6 0

3 years ago