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2 years ago

Is there a trick to find?

Mathematics

2 answers:

Studentka2010 [4]2 years ago

6 0

Maybe simplifying it otherwise maybe not

tigry1 [53]2 years ago

4 0

Answer:

I don't believe so

Step-by-step explanation:

generally for fractions, you can start by simplifying

to simplify divide the numerator and denominator by a common factor

in this case, there is no common factor, so this either your answer or you can divide 729 by 4096 to get a decimal number (0.118)

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5y^2-6y-11-6y^2+2y+5

shusha [124]

Answer:

-y² - 4y - 6

Step-by-step explanation:

Step 1: Write expression

5y² - 6y - 11 - 6y² + 2y + 5

Step 2: Combine like terms (y²)

-y² - 6y - 11 + 2y + 5

Step 3: Combine like terms (y)

-y² - 4y - 11 + 5

Step 4: Combine like terms (constants)

-y² - 4y - 6

6 0

2 years ago

SOMEONE HELP ASAPPPP PLEASEEE,PLEASE EXPLAIN HOW U GOT YOUR ANSWER, I NEED AN EXPLANATION IN ORDER TO COMPLETE THIS! NO LINKS OR

artcher [175]

Answer:

ok so just do your best all you need to do is (x,y) (y,x) or make them negitive so just do ur best i believe in you ok you got this

Step-by-step explanation:

4 0

2 years ago

Suppose a circle is divided into congruent slices. Match each number of slices with the

Stells [14]

26 skate and with wunno when ask what about the gc and the park to park in

8 0

2 years ago

PLEASE HELLPPPP ASSSAPPP

Vera_Pavlovna [14]

There’s an app called photo math where u can take a picture of the problem and it gives u the answer.

hope that helps lol

4 0

2 years ago

What number must you add to complete the square?<br> x^2 + 2x = -1

Andru [333]

$\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}$

If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}$

6 0

2 years ago

Read 2 more answers

Is there a trick to find?​

Is there a trick to find?