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2 years ago

How to add 3/5 and 1/10 in simplest form

Mathematics

2 answers:

dolphi86 [110]2 years ago

5 0

Find a common denominator and then add.
3/5=6/10
6/10+1/10=7/10

Mekhanik [1.2K]2 years ago

4 0

To add these you must have a common denominator. In this case 10 will work because both 5 and 10 can evenly go into 10. Now whatever you multiplied the bottom by to get 10 you must do the same to the top. This is shown below:
3/5+ 1/10 = 6/10+ 1/10
You can now add these together to get 7/10. This is also in the simplest form.

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Need help on 5. 6. And 7. 7.c is (6+x)-8 7.d is 8-(6+x)

enyata [817]

#7 is A
For # 5 & 6 is there a multiplication sign between the numbers or is supposed to be a power or something else?

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3 years ago

An electrician charges a flat service fee to accept a job and an additional hourly rate to complete the job. The amount an elect

Colt1911 [192]

The interpretation of the slope in the context of the problem is: For every hour the electrician works, her pay will increase $28.

<h3>How to Interpret the Slope of a Function?</h3>

The slope of any function is the unit rate or average rate of change of the given function for that situation.

Given the function, E(x) = 130 + 28x, where:

E(x) is the amount an electrician charges per job

130 is the flat rate service which is the starting value or y-intercept

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Thus, the interpretation of the slope in the context of the problem is: For every hour the electrician works, her pay will increase $28.

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brainly.com/question/24502219

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2 years ago

A beehive contains 5 gallons of honey a beekeeper removes 4 gallons one pint and 2 oz how much honey remains in the beehive

sergejj [24]

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2 years ago

Which answer best describes the complex zeros of the polynomial function?

marysya [2.9K]

Answer-

D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.

Solution-

The given polynomial is,

$f(x)=x^3-x^2+6x-6$

The zeros of the polynomials are,

$\Rightarrow f(x)=0$

$\Rightarrow x^3-x^2+6x-6=0$

$\Rightarrow x^2(x-1)+6(x-1)=0$

$\Rightarrow (x^2+6)(x-1)=0$

$\Rightarrow (x^2+6)=0,(x-1)=0$

$\Rightarrow x=\sqrt{-6},\ x=1$

$\Rightarrow x=1,\ \pm \sqrt6i$

Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1

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3 years ago

Help me with the question of o.math

Lelu [443]

Answer:

Hello,

Answer 163

Step-by-step explanation:

1,7,16,28,43,..., 10th term=?

$u_1=1,u_2=7,u_2-u_1=7-1=6=2*3\\u_3=16, u_3-u_2=16-7=9=3*3\\u_4=28, u_4-u_3=28-16=12=4*3\\u_5=43, u_5-u_4=43-28=15=5*3\\\\\boxed{ u_1=1, \ u_{n+1}-u_{n}=(n+1)*3}\\\\u_{10}=163\\$

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7 0

3 years ago

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