What is 15 3/5 % in simplest form

Idk if she's too young to learn cross multiply, but this is an easier method.

Multiply everything by 4 which cancels out the 4 on the left side.
36z-16/5 <= 7z + 2
Now multiply everything by 5, cancelling out the right side 5.
36z-16 <= 35z + 10
Solve the equation normally
z-16<=10

z<=36

3 0

2 years ago

A rectangular prism with a length of 7 centimeters, width of 5 centimeters, and height of 8 centimeters. A rectangular prism wit

Genrish500 [490]

Answer: The volume of the composite figure is $420\text{ cm}^3$ .

Step-by-step explanation:

We know that the volume of a rectanbgular prism is given by :-

Volume = length x width x height

Given, for prism 1 , we have

length of 7 centimeters, width of 5 centimeters, and height of 8 centimeters

Volume of prism 1= $(7)(5)(8)=280\text{ cm}^3$

Given, for prism 2 , we have

length of 10 centimeters, width of 7 centimeters, and height of 4 centimeters.

Volume of prism 2= $(10)(7)(4)=280\text{ cm}^3$

Now , the volume of the composite figure = Volume of prism 1+ Volume of prism 2

$=210\text{ cm}^3+210\text{ cm}^3=420\text{ cm}^3$

Hence, the volume of the composite figure is $420\text{ cm}^3$ .

5 0

2 years ago

Read 2 more answers

What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

4 0

1 year ago