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2 years ago

Write the fraction in lowest terms. 13/17

Mathematics

2 answers:

Natasha2012 [34]2 years ago

5 0

Those are both prime numbers so so there the lowest

Dmitry [639]2 years ago

3 0

13/17 is at the lowest because they are both prime numbers.

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The March 11, 2011, earthquake in Japan caused a tsunami that crashed into the (cant say on here because it looks like a bad wor

Lesechka [4]

A radioactive atmosphere around the area and an expensive, hard time cleaning everything up.

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2 years ago

Please help 15 points<br> x2≥−4

Nutka1998 [239]

Answer:

Solve for

by simplifying both sides of the inequality, then isolating the variable.

Step-by-step explanation:

6 0

3 years ago

Unit 7:Polygons & Quadrilaterals Homework 4: Rhombi and squares

iogann1982 [59]

Based on the quadrilateral given, the values of the missing measures will be:

VU = 15
SU = 21.21
TV = 21.21
SW = 10.61

<h3>Solving quadrilateral.</h3>

A quadrilateral simply means a polygon that has four sides, edges and vertices.

Based on the information given, SU = TV and this will be:

= ✓450 = 21.21

SW = 21.21/2 = 10.61

Also, for the second shape, the values will be:

OM = 46; PN = 23, ON = 46; MN = 46.

Learn more about quadrilateral on:

brainly.com/question/224113

4 0

2 years ago

During the first week of a summer camp 2 out of 3 campers were boys. During the second week 3 out of 5 were boys. There were a 1

rewona [7]

Step-by-step explanation:

2/3=10/15 (divide 15 by 3 to get 5, then multiply both numerator and denominator by 5 to set equivalent)

3/5=9/15

10/15 is greater than 9/15, so during the 1st week there were more boy campers

6 0

3 years ago

Find dy/dx using first principle<br>x+1/x

JulsSmile [24]

Answer:

The derivative of the function is:

$f'(x)=1-\frac{1}{x^{2}}$

Step-by-step explanation:

The first principle is given by:

$f'(x)=\frac{dy}{dx}=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The function here is:

$f(x)=x+\frac{1}{x}$

Now, using the first principle we have:

$f'(x)=lim_{h\rightarrow 0}\frac{(x+h)+\frac{1}{(x+h)}-x-\frac{1}{x}}{h}$

$=lim_{h\rightarrow 0}\frac{h+\frac{1}{(x+h)}-\frac{1}{x}}{h}$

$=lim_{h\rightarrow 0}\frac{h-\frac{h}{x(x+h)}}{h}$

$=lim_{h\rightarrow 0}\frac{h(1-\frac{1}{x(x+h)})}{h}$

$=lim_{h\rightarrow 0}1-\frac{1}{x(x+h)}$

$=1-\frac{1}{x^{2}}$

Therefore, the derivative of the function is:

$f'(x)=1-\frac{1}{x^{2}}$

I hope it helps you!

6 0

2 years ago