All categories

3 years ago

What is the solution to this system of equations?

Mathematics

1 answer:

lesya692 [45]3 years ago

7 0

Answer:

The solution to this system of equations

Step-by-step explanation:

D) infinitely many solutions

You might be interested in

7+x+7x equivalent expressions

bearhunter [10]

8x+7, 7x+x+7, 7+x(1+7)

8 0

4 years ago

Find the value of each variable in each parallelogram.

svp [43]

Answer:

a=109° b=71°

Step-by-step explanation:

All the angles in a parallelogram add up to 360. There are a few rules:

The opposite angles are congruent.

Consecutive angles are congruent.

We can apply these rules here.

71 is opposite to b

and a is consecutive to 71 so we can put together these equations.

180=71+a

71=b

we can simplify to get the answers

the end

4 0

3 years ago

6xy (x 2 - xy + y 2 )

Liono4ka [1.6K]

(if 2 is a square)

6xy ( x2 - xy + y2)

= 6x3y - 6x2y2 - 6xy3 --------------the threes and twos on this line are squares and cubics. the six is just a whole number

6 0

4 years ago

Anyone please !!!!!!

IgorLugansk [536]

Answer:

Step-by-step explanation:

Lets start by plugging in our (a) value

a^2 + 5a + 4

(-5)^2 + 5(-5) +4

25 - 25 + 4

0 + 4

4 is your final answer!

5 0

3 years ago

A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0

3 years ago