All categories

3 years ago

Find each sum or difference -8/35+2/5=

Mathematics

2 answers:

Alexxandr [17]3 years ago

8 0

Answer:

6/35

Step-by-step explanation:

change 2/5 to 14/35 so the denominators are even.

after that you get -8/35+14/35

then you get 6/35

Lena [83]3 years ago

6 0

First find the common denominator.

-8/35 + (2)(7)/(5)(7) = -8/35 + 14/35

-8/35 + 14/35 = 6/35

Answer: 6/35

You might be interested in

Please help me with this question!

Westkost [7]

Answer:

Your Answer Is 10

Step-by-step explanation:

1st Factor the Numerator and Denominator then cancel the common factor.

Hope It Help You If Yes then Please make me the Brainiest.

Thank You

Parshv

7 0

3 years ago

What is the product of 2x + y and 5x – y + 3?

Radda [10]

Answer:

The product of 2x + y and 5x – y + 3 is 10x^2+ 3xy + 6x - y^2 + 3y

Step-by-step explanation:

The product of the two expressions can be expressed mathematically as;

(2x +y) (5x -y +3)

To obtain the product of these expressions, we simply expand the brackets as follows;

2x(5x -y +3) + y(5x -y +3)

= 10x^2 - 2xy + 6x + 5xy - y^2 + 3y

= 10x^2- 2xy + 5xy + 6x - y^2 + 3y

= 10x^2+ 3xy + 6x - y^2 + 3y

The product of 2x + y and 5x – y + 3 is thus 10x^2+ 3xy + 6x - y^2 + 3y

5 0

3 years ago

Read 2 more answers

2. Sophia scored 42 points in 3 games. How many points would you expect her to make in 11 games?

igomit [66]

Answer:

154

Step-by-step explanation:

42/3=14

14*11=154

6 0

2 years ago

Read 2 more answers

I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

3 0

3 years ago

Do you line plots display individual data

tino4ka555 [31]

Yes they do, in the grand aspect of things.

5 0

3 years ago