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SVETLANKA909090 [29]
3 years ago
9

PLEASE FREAKING HELP ME ASAP OMG I REALLY NEED HELP ITS DUE IN 8 MINS PLS PLS EXPLAIN UR ANSWER STEP BY STEP HURRY PLSSSSSSSSSSS

SSSSSSS

Mathematics
1 answer:
Blababa [14]3 years ago
6 0

Answer:

396m=396m.

your answer is 396m

hope this helps :)

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Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments
kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

7 0
1 year ago
Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1<br>​
slamgirl [31]

Answer:Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1

Step-by-step explanation:

3 0
3 years ago
hi i'm really struggling with this and its due later today! If someone could show the work of how to do it and the answer that w
GuDViN [60]

Answer:

\frac{-5 \pm \sqrt{13} }{6}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\frac{-5 \pm \sqrt{5^2-4(3)(1)} }{2(3)}<u />

<u />

<u>Step 2: Evaluate</u>

  1. Evaluate Exponents:                    \frac{-5 \pm \sqrt{25-4(3)(1)} }{2(3)}
  2. Evaluate Multiplication:               \frac{-5 \pm \sqrt{25-12} }{6}
  3. Evaluate Subtraction:                  \frac{-5 \pm \sqrt{13} }{6}
8 0
3 years ago
Divide and write your answer in simplest form.<br> 4 ÷ 1/2
MAVERICK [17]

Answer:

the answer will be 8

Step-by-step explanation:

pls give me brainliest

8 0
3 years ago
Write the equation in standard form for the circle that has a diameter with endpoints (- 1, 12) and (- 1, 8)​
matrenka [14]

Answer:

The equation in standard form is (x + 1)^{2} + (y - 10)^{2} = 4

Step-by-step explanation:

The distance from (-1, 12) to (-1, 8) = 4 = diameter of the circle.  So, 2 = radius

The center of the circle is at (-1, 10)

The equation in standard form is (x + 1)^{2} + (y - 10)^{2} = 4

3 0
3 years ago
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