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kumpel [21]
3 years ago
7

Quadrilateral ABCD has the following

Mathematics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

No because the angle of point C is NOT congruent to the angle of point A.

Step-by-step explanation:

A quadrilateral MUST be a parallelogram if it has both pairs of its opposite angles congruent.

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Divide 2x2+7x-3 by 2x+5. Which expression represents the quotient and remainder
Fiesta28 [93]

Answer:

0

Step-by-step explanation:

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Since 1952 the total number of people living in wales has increased by about one eighth.
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3 0
3 years ago
For what value of k, the zeroes of x2 + kx + 12 will differ by 1?
asambeis [7]

Let <em>a</em> and <em>b</em> be the zeroes of <em>x</em>² + <em>kx</em> + 12 such that |<em>a</em> - <em>b</em>| = 1.

By the factor theorem, we can write the quadratic in terms of its zeroes as

<em>x</em>² + <em>kx</em> + 12 = (<em>x</em> - <em>a</em>) (<em>x</em> - <em>b</em>)

Expand the right side and equate the coefficients:

<em>x</em>² + <em>kx</em> + 12 = <em>x</em>² - (<em>a</em> + <em>b</em>) <em>x</em> + <em>ab</em>

Then

<em>a</em> + <em>b</em> = -<em>k</em>

<em>ab</em> = 12

The condition that |<em>a</em> - <em>b</em>| = 1 has two cases, so without loss of generality assume <em>a</em> > <em>b</em>, so that |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>.

Then if <em>a</em> - <em>b</em> = 1, we get <em>b</em> = <em>a</em> - 1. Substitute this into the equations above and solve for <em>k</em> :

<em>a</em> + (<em>a</em> - 1) = -<em>k</em>   →   2<em>a</em> = 1 - <em>k</em>   →   <em>a</em> = (1 - <em>k</em>)/2

<em>a</em> (<em>a</em> - 1) = 12   →   (1 - <em>k</em>)/2 • ((1 - <em>k</em>)/2 - 1) = 12

→   (1 - <em>k</em>)²/4 - (1 - <em>k</em>)/2 = 12

→   (1 - <em>k</em>)² - 2 (1 - <em>k</em>) = 48

→   (1 - 2<em>k</em> + <em>k</em>²) - 2 (1 - <em>k</em>) = 48

→   <em>k</em>² - 1 = 48

→   <em>k</em>² = 49

→   <em>k</em> = ± √(49) = ±7

8 0
3 years ago
Need help with this one please! Worth 30 points!!! :)
Kitty [74]

Answer:

y^2 +8y + 16

Step-by-step explanation:

 6y^2 +2y +5  - (5y^2 -6y -11)

I distribute the minus sign

6y^2 +2y +5  - 5y^2 +6y +11

Then I put them vertical.

6y^2 +2y +5  

-5y^2 +6y +11

------------------------

y^2 +8y + 16


This is in standard from since the exponential decreases.

5 0
3 years ago
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