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valentinak56 [21]
2 years ago
6

Given A(2, 3) B(1, 4) C(1, 3) D(-2, 2). How are AB and CD related?

Mathematics
1 answer:
kondaur [170]2 years ago
3 0

The vector AB is not related with the vector CD as k is not the same for each pair of components.

<h3>Are two vectors similar?</h3>

In this question we must prove if the vector AB is a multiple of the vector CD, that is:

\overrightarrow{AB} = k \cdot \overrightarrow {CD}

\vec B - \vec A = k \cdot [\vec D - \vec C]

(1, 4) - (2, 3) = k · [(- 2, 2) - (1, 3)]

(- 1, 1) = k · (- 3, - 1)

Hence, the vector AB is not related with the vector CD as k is not the same for each pair of components.

To learn more on vectors: brainly.com/question/13322477

#SPJ1

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The circle graph shows the results of a survey in which 250 students were asked to choose their favorite color. How many more st
Alex Ar [27]

Answer:

The correct answer is 4%

Step-by-step explanation:

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3 years ago
For which value of x must the square root <br><br> 97x<br><br> be further simplified
lys-0071 [83]

Question:

For which value of x must the expression √97x be further simplified?

A. 5    B. 10     C. 15     D. 20

Answer:

D.  x = 20

Step-by-step explanation:

Given

\sqrt{97x

Required

For what value of x can it be simplified

From the list of options, only (D) is correct

D. 20

Substitute 20 for x in \sqrt{97x

\sqrt{97 x} = \sqrt{97 * 20}

Express 20 as 4 * 5

\sqrt{97 x} = \sqrt{97 * 4 * 5}

Reorder

\sqrt{97 x} = \sqrt{4 *97 * 5}

Split

\sqrt{97 x} = \sqrt{4} *\sqrt{97 * 5}

\sqrt{97 x} = 2*\sqrt{97 * 5}

\sqrt{97 x} = 2*\sqrt{485}

\sqrt{97 x} = 2\sqrt{485}

With other values of x, the given expression can not be simplified.

Hence:

x = 20

6 0
3 years ago
Solve the following:<br> a. 4x^2 + 5x + 3 = 2x^2 − 3x<br> b. c^2 − 14 = 5c
Jet001 [13]

Answer:

(a) x = -0.418 , -3.581

(B) c = 6.855, -1.855

Step-by-step explanation:

(A) We have given equation 4x^2+5x+3=2x^2-3x

2x^2+8x+3=0

On comparing with standard quadratic equation ax^2+bx+c

a = 2, b = 8 and c = 3

So roots of the equation will be x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-8\pm \sqrt{8^2-4\times 2\times 3}}{2\times 2}=\frac{-8\pm 6.324}{4}=-0.418,-3.581

(b) c^2-14=5c

c^2-5c-14=0

a = 1, b = -5 and c= -14

So c=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm \sqrt{(-5)^2-4\times 1\times (-14)}}{2\times 1}=\frac{5\pm 8.71}{2}=6.855,-1.855

7 0
3 years ago
Which side of the dilation corresponds to side AC of the original figure?
Sav [38]
It is A'C' because when the figure is dilatted the corresponding figure has " by it.

hopes this helps :)

8 0
3 years ago
A system of equations is shown below.
Nimfa-mama [501]
-3x + 7y = -16 . . . (1)
-9x + 5y = 16 . . . .(2)

(1) x 3 => -9x + 21y = -48 . . . (3)
(2) - (3) => -16y = 64 . . . (4)

Required equivalent system of equations is
-9x + 5y = 16
-16y = 64
5 0
3 years ago
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