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ki77a [65]
3 years ago
10

How to solve this problem step by step. Make brainliest answer if its done. Please

Mathematics
1 answer:
sladkih [1.3K]3 years ago
6 0

D is halfway between A and B

so the coordinates of D are (2,2)

E is halfway between A and C so the coordinates of E are (-1,1)

now you need to find the gradient/slope of DE and BC using the formula:

\frac{y2 - y1}{x2 - x1}

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>D</u><u>E</u><u>:</u><u> </u></h3>

SUB IN COORDINATES OF D AND E

\frac{1 - 2}{ - 1 - 2}

therefore the gradient of DE is 1/3.

<h3><u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>B</u><u>C</u><u>:</u></h3>

<em>S</em><em>U</em><em>B</em><em> </em><em>I</em><em>N</em><em> </em><em>C</em><em>O</em><em>O</em><em>R</em><em>D</em><em>I</em><em>N</em><em>A</em><em>T</em><em>E</em><em>S</em><em> </em><em>O</em><em>F</em><em> </em><em>B</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>C</em>

<em>\frac{ - 2 - 0}{ - 3 - 3}</em>

therefore the gradient of BC is -2/-6 which simplifies to 1/3.

<h3>therefore, BC and DE are parallel as they both have a gradient/slope of 1/3 and parallel lines have the same gradient</h3>

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(50 points) need help.
tatiyna

Answer:

I believe it is 127ft²

Step-by-step explanation:

6 0
2 years ago
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For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.
andriy [413]

Answer:

If k = 1 or k = -\frac{1}{2}, there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

The signal of \bigtriangleup determines how many real roots an equation has:

\bigtriangleup > 0: Two real and different solutions

\bigtriangleup = 0: One real solution

\bigtriangleup < 0: No real solutions

In this problem, we have the following second order polynomial:

(k+1)x^{2} + 4kx + 2 = 0.

This means that a = k+1; b = 4k; c = 2

It has one solution if

\bigtriangleup = 0

b^{2} - 4ac = 0

16k^{2} -8(k+1) = 0

16k^{2} - 8k - 8 = 0

We can simplify by 8

2k^{2} - k - 1 = 0

The solution is:

k = 1 or k = -\frac{1}{2}

So, if k = 1 or k = -\frac{1}{2}, there is only one solution to the given quadratic equation.

4 0
3 years ago
What is the value of x in the equation 3x4y=65, when y=4
bixtya [17]

Answer:

The value of x if the equation is

3x + 4y=65

  x = 49/3 or 16.3

&

3x - 4y=65

   x= 27

5 0
3 years ago
Read 2 more answers
16,42 greatest common factor
kozerog [31]
2

42 ÷ 2 = 21
16 ÷ 2 = 8

Let's try the other factors of 42...

42 ÷ 6 = 7
16 ÷ 6 = 2.66
No...

42 ÷ 7 = 6
16 ÷ 7 = 2 2/7
No...

42 ÷ 14 = 3
16 ÷ 14 = 1.1428
No...
7 0
3 years ago
Can i have help on this q plz
Alexandra [31]

Answer:

14.8

Step-by-step explanation:

7.4 * 2 = 14.8.

hope this helps

8 0
2 years ago
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