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andriy [413]
3 years ago
9

POINT out the correct ANSWER (Free_

Mathematics
2 answers:
svlad2 [7]3 years ago
4 0

Answer:

40...................................

lina2011 [118]3 years ago
4 0

Answer:

35 points because you said so O-O

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What does the distributive property combine to make multiplying whole numbers easier?
IgorC [24]
An algebra property whuch is used to multiply a single term and two or more terms inside a set of parentheses
5 0
4 years ago
What is the product? -4. [8/-1/-5/9]
Bad White [126]

Answer:

- 32, 4, 20, and - 36

Step-by-step explanation:

We have to find the product of -4 with 8 / - 1 / - 5 / 9.

Now, the product of - 4 and 8 is = - 32  

{Since the product of a positive number and a negative number is always negative}

The product of - 4 and -1 is = 4

{Since the product of two negative numbers is always positive}

The product of - 4 and - 5 is = 20

The product of - 4 and 9 is = -36

(Answer)

8 0
3 years ago
An equivalent expression to a2/3
scoundrel [369]

Answer:

4/6  6/9  8/12  10/15 and so on, just multiply both the numerator and denominator by the same number and get your answer.

Step-by-step explanation:

6 0
4 years ago
Can someone help me answer this?
andriy [413]

Answer:

m<S = 68 degrees.

m<D = 112 degrees

Step-by-step explanation:

Opposite angles of a parallelogram are equal so:

4x - 4 = 3x + 14      

4x - 3x = 14 + 4

x = 18.

So m < S = 4(18) - 4

= 72-4

= 68 degrees.

Angles on the same side of a parallelogram are supplementary, so

m < D = 180 - 68

= 112 degrees.

7 0
2 years ago
What is the perimeter of this quadrilateral?<br> (5,5)<br> (2, 4)<br> (4, 1)<br> (6, 1)
lbvjy [14]

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}

BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}

DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89

8 0
3 years ago
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