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3 years ago

Why do you first combine like terms and then use Distributive Property to solve 8(5x+9x+6)=160?

1 answer:

5 0

You can actually do it either way, you could distribute first and then combine, you would get the same answer. I guess it just depends on what you find easier. Here is how to solve it both ways:

Combining First:

8(5x+9x+6)=160

8(14x+6)=160

112x+48=160

112x=112

x=1

NOT Combining First:

8(5x+9x+6)=160

40x+72x+48=160

112x+48=160

112x=112

x=1

so, you would get the same answer both ways, it is of personal preference what you do first.

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Anyone understand this? i need help :((

Zinaida [17]

Answer:

(-9,8)

(-2,8)

(-1,8)

(0,8)

(7,8)

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2 years ago

Geometry help needed!

salantis [7]

Answer:

AC Midpoint (2,0)-

CB Midpoint (5,-4)

midlength is 5

Step-by-step explanation:

midpoint formula $(\frac{x_{1}+x_{2} }{2} ,\frac{y_{1} +y_{2} }{2})$

A(2,4) C(2,-4)

$\frac{2+2}{2} =\frac{4}{2}=2\\\frac{4-4}{2} =\frac{0}{2} =0$

(2,0)

C(2,-4) B(8,-4)

$\frac{8+2}{2} =\frac{10}{2} =5$

$\frac{-4-4}{2} =\frac{-8}{2} =-4$ midpoint is (5,-4)

Distance between the 2 midpoints is the midsegment

(2,0) and (5,-4)

$d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1}) ^{2} }$

$d=\sqrt{(5-2)^{2} +(-4-0)^{2} } \\d=\sqrt{3^{2} +(-4^{2})\\ } \\d=\sqrt{x} 9+16\\d=\sqrt{25} \\d=5$

3 0

3 years ago

Write an equation of the line below.

statuscvo [17]

Answer:

y = -4x - 4

Step-by-step explanation:

Rise/Run

-8/2 = -4

Y-intercept:

-4

8 0

2 years ago

David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o

BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :( $b^4,b^2,b^4$ )

$b^4$ = b × b × b × b

$b^2$ = b × b

$b^4$ =b × b × b × b

GCF of variables :( $b^4,b^2,b^4$ ) is $b^2$

3) GCF of $c^3$ and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.