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Musya8 [376]
1 year ago
5

Expand and simplify: (-7y - 8) (y - 4)

Mathematics
1 answer:
ZanzabumX [31]1 year ago
8 0
<h2><u>Simplification</u></h2>

<h3>Expand and simplify: (-7y - 8) (y - 4)</h3>

To expand an expression, use FOIL method.

<h3><u>F - First</u></h3>

  • Multiply the first term in each expression inside the parentheses.

<u>-7y</u> and <u>y</u> are the first terms in <u>each expression</u>.

<u>(-7y)(y) = -7y²</u>

<h3><u>O - Outside </u></h3>

  • Next is to multiply the terms outside. (The first term in the first expression and the last term in the second expression)

<u>-7y</u> is the first term in the <u>first expression</u>

<u>-4</u> is the last term of the <u>second expression</u>

<u>(-7y)(-4) = 28y</u>

<h3><u>I - Inside</u></h3>

  • The next step is to multiply the terms inside. (The last term in the first expression and the first term in the last expression)

<u>-8</u> is the last term in the <u>first expression</u>

<u>y</u> is the first term in the <u>second expression</u>

<u>(-8)(y) = -8y</u>

<h3><u>L - Last</u></h3>

  • Last step is to multiply the last term in each expression.

<u>-8</u> is the last term of the <u>first expression</u>

<u>-4</u> is the last term of the <u>second expression</u>

<u>(-8)(-4) = 32</u>

Combine all simplified terms.

  • -7y² + 28y - 8y + 32
  • -7y + 20y + 32

<u>Answer:</u>

  • <u>-7y + 20y + 32</u>

Wxndy~~

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Answer:

(3,2)

Step-by-step explanation:

Let's solve this via elimination method:

\begin{cases} x+y=5\\x-y=1\end{cases}

By subtracting equation 2 from equation one we obtain:

\begin{cases} x+y=5\\-(x-y=1)\end{cases}\\\\0x+2y=4\\\\y=2

Next we can use any equation either 1 or 2 to determine what x is, I'll use equation 1.  Let y=2 and so:

x+y=5\\\\x+2=5\\x=5-2\\\\x=3

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3 years ago
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Serggg [28]
37.5% of 32 is 12

if you move the decimal point in the percentage 2 places to the left you will get a number you can multiply with 

so 0.375 x 32 = 12


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3 years ago
Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real
hoa [83]

Answer: 0=2x^2-7x-9\\0=4x^2-3x-1\\0=x^2-2x-8


Step-by-step explanation:

We know that the standard quadratic equation is  ax^2+bx+c=0

Let's compare all the given equation to it and , find discriminant.

1. a=2, b= -7, c=-9

b^2-4ac=(-7)^2-4(2)(-9)=49+72>0

So it has 2 real number solutions.

2. a=1, b=-4, c=4

b^2-4ac=(-4)^2-4(1)(4)=16-16=0

So it has only 1 real number solution.

3. a=4, b=-3, c=-1

b^2-4ac=(-3)^2-4(4)(-1)=9+16=25>0

So it has 2 real number solutions.

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b^2-4ac=(-2)^2-4(1)(-8)=4+32=36>0

So it has 2 real number solutions.

5. a=3, b=5, c=3

b^2-4ac=(5)^2-4(3)(3)=25-36=-9

Thus it does not has real solutions.



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