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2 years ago

Use the Distributive Property to simplify the expression. [Example: 5(4n-3) = 20n - 15] 7(5n-9)

Mathematics

1 answer:

pochemuha2 years ago

7 0

Answer:

I got 17n=5n]26n

I don't know what to do next but I hope it helps ^^

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How do you simplify 27 over 20

soldi70 [24.7K]

Answer:

1 7/20

Step-by-step explanation:

well 27/20 is 1 and 7/20

make it to a improper fraction

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2 years ago

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The price of a new SMART television was reduced from $250 to $200. By what percentage was the price of the television reduced?

Novosadov [1.4K]

Answer:

20%

Step-by-step explanation:

if the 100% of the price is $250 and the discount is $50 then to find what percent $50 we multiply 100 by 50 then divide it with 250

100 × 50 ÷ 250 = 20

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2 years ago

Quadrilateral JKLM is graphed with vertices at J(-2,2), K(-1,-5), L(4,0), and M(3,7).

AveGali [126]

Answer:

<u>Question (a)</u>

Midpoint of a line segment:

$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Given:

J = (-2, 2)
L = (4, 0)

$\implies \textsf{midpoint of }JL=\left(\dfrac{-2+4}{2},\dfrac{2+0}{2}\right)=(1,1)$

Given:

M = (3, 7)
K = (-1, -5)

$\implies \textsf{midpoint of }MK=\left(\dfrac{3-1}{2},\dfrac{7-5}{2}\right)=(1, 1)$

<u>Question (b)</u>

Find slopes (gradients) of JL and MK then compare. If the product of the slopes of JL and MK equal -1, then JL and MK are perpendicular.

$\textsf{slope }m=\dfrac{y_2-y_1}{x_2-x_1}$

Given:

J = (-2, 2)
L = (4, 0)

$\implies \textsf{slope of }JL=\dfrac{0-2}{4+2}=-\dfrac13$

Given:

M = (3, 7)
K = (-1, -5)

$\implies \textsf{slope of }MK=\dfrac{-5-7}{-1-3}=3$

$\textsf{slope of }JL \times \textsf{slope of }MK=-\dfrac13 \times3=-1$

Hence segments JL and MK are perpendicular

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1 year ago

NEED HELP ON THIS QUESTION ASAP

kotykmax [81]

Answer:

It could be B

Step-by-step explanation:

5 0

2 years ago

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If two polynomial equations have real solutions, then will the equation that is the result of adding, subtracting, or multiplyin

kirill115 [55]

No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.

Consider polynomial equations x^2 = 3 and x^2 = 1.

Obviously they have real solutions.

Subtract the two polynomial equations:

(x^2 - x^2) = (3 - 1)

0 = 2...

We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.

Obviously 0 = 2 has no real solution.

3 0

2 years ago