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sveticcg [70]
3 years ago
8

whats another way to write (s-6)(s+1) when s represents the side of a square and the equation represents the area of a rectangle

?
Mathematics
1 answer:
viktelen [127]3 years ago
8 0

The another way to write (s - 6)(s + 1) is s² - 5s - 6

Step-by-step explanation:

Let us revise how multiply two binomial (a + b)(x + y)

  • Multiply the 1st terms
  • Multiply the 2nd terms
  • Multiply the nears and extremes, where nears are the 2nd term of first bracket and 1st term in the second bracket, the extremes are the 1st term in the first bracket and the 2nd term in the second bracket
  • Add the like terms if necessary

∵ The area of the square = (s - 6)(s + 1)

- To find the another way multiply the two brackets

∵ s × s = s²

∵ -6 × 1 = -6

∵ -6 × s = - 6s ⇒ nears

∵ s × 1 = s ⇒ extremes

- The terms - 6s and s are like terms, then add them

∵ - 6s + s = - 5s

∴ (s - 6)(s + 1) = s² - 5s - 6

∴ The area of the square = s² - 5s - 6

The another way to write (s - 6)(s + 1) is s² - 5s - 6

Learn more:

You can learn more about the binomials in brainly.com/question/2334388

#LearnwithBrainly

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Step-by-step explanation:

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you give 1/3 of a pan of brownies to Susan a 1/6 of the pan of brownies to Patrick. how much of the pan of brownies did you give
Neko [114]
5/10 is the answer or 1/2 if reduced
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3 years ago
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Lisa has a certain amount of money. She spent 39 dollars and has 3/4th of the original amount left. How much money did she have
harina [27]
If \frac { 3 }{ 4 } is left, that means \frac { 1 }{ 4 } was spent. So \frac { 1 }{ 4 } of the total amount is equal to 39.. Let's say x to the total amount. \frac { 1 }{ 4 } \cdot x=39\\ x=4\cdot 39\\ x=156\\

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3 years ago
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For which system of equations is (5, 3) the solution? A. 3x – 2y = 9 3x + 2y = 14 B. x – y = –2 4x – 3y = 11 C. –2x – y = –13 x
Alla [95]
The <u>correct answer</u> is:

D) \left \{ {{2x-y=7} \atop {2x+7y=31}} \right..

Explanation:

We solve each system to find the correct answer.

<u>For A:</u>
\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.

Since we have the coefficients of both variables the same, we will use <u>elimination </u>to solve this.  

Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
\left \{ {{3x-2y=9} \atop {+(3x+2y=14)}} \right. &#10;\\&#10;\\6x=23

Next we divide both sides by 6:
6x/6 = 23/6
x = 23/6

This is <u>not the x-coordinate</u> of the answer we are looking for, so <u>A is not correct</u>.

<u>For B</u>:
\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.

For this equation, it will be easier to isolate a variable and use <u>substitution</u>, since the coefficient of both x and y in the first equation is 1:
x-y=-2

Add y to both sides:
x-y+y=-2+y
x=-2+y

We now substitute this in place of x in the second equation:
4x-3y=11
4(-2+y)-3y=11

Using the distributive property, we have:
4(-2)+4(y)-3y=11
-8+4y-3y=11

Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
y=19

This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>B is not correct</u>.

<u>For C</u>:
Since the coefficient of x in the second equation is 1, we will use <u>substitution</u> again.

x+2y=-11

To isolate x, subtract 2y from each side:
x+2y-2y=-11-2y
x=-11-2y

Now substitute this in place of x in the first equation:
-2x-y=-13
-2(-11-2y)-y=-13

Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
22+4y-y=-13

Combining like terms:
22+3y=-13

Subtract 22 from each side:
22+3y-22=-13-22
3y=-35

Divide both sides by 3:
3y/3 = -35/3
y = -35/3

This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>C is not correct</u>.  

<u>For D</u>:
Since the coefficients of x are the same in each equation, we will use <u>elimination</u>.  We have 2x in each equation; to eliminate this, we will subtract, since 2x-2x=0:

\left \{ {{2x-y=7} \atop {-(2x+7y=31)}} \right. &#10;\\&#10;\\-8y=-24

Divide both sides by -8:
-8y/-8 = -24/-8
y=3

The y-coordinate is correct; next we check the x-coordinate  Substitute the value for y into the first equation:
2x-y=7
2x-3=7

Add 3 to each side:
2x-3+3=7+3
2x=10

Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so <u>D is the correct answer</u>.
7 0
3 years ago
If the factors of function f are (x - 6) and (x - 1), what are the zeros of function f?
lidiya [134]
X=6 and x=1

You can find your zeros by determining what you have to plug into the function in order for it to equal zero

If we plug in 6, for example we’d get (6-6)(x-1)

Simplified this is 0(x-1)

Anything times 0 is 0, so this is one of our zeros.

Same goes for x-1, we just need to plug in 1 for it to equal 0

Therefore there are zeros at x=1 and x=6 :))
7 0
3 years ago
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