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e-lub [12.9K]
3 years ago
8

20 points

Mathematics
1 answer:
irinina [24]3 years ago
4 0

Answer:

(a) 9x^2+16x-4

(b) 9a(2a+5)

(c) (y-6)(y-4)

Step-by-step explanation:

(a)

(9x-2)(x+2)=

9x^2-2x+18x-4=

9x^2+16x-4

(b)

18a^2+45a=

9a(2a+5)

(c)

y^2-10y+24=

(y-6)(y-4)

Hope this helps!

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Solve the equation: k^2+5k+13=0
mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

x =  \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}  \\

From the question

a = 1 , b = 5 , c = 13

So we have

k =  \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)}  \\  =  \frac{ - 5 \pm \sqrt{25 - 52} }{2}  \\  =  \frac{ - 5 \pm \sqrt{ - 27} }{2}  \:  \:  \:  \:  \:  \:  \\  =  \frac{ - 5  \pm3 \sqrt{3}  \: i}{2}  \:  \:  \:  \:  \:  \:

<u>Separate the solutions</u>

k_1 =  \frac{ - 5 + 3 \sqrt{3} \: i }{2}  \:  \:  \:  \: or \\ k_2 =  \frac{ - 5 - 3 \sqrt{3}  \: i}{2}

The equation has complex roots

<u>Separate the real and imaginary parts</u>

We have the final answer as

k_1 =  -  \frac{5}{2}  +  \frac{3 \sqrt{3} }{2}  \: i \:  \:  \:  \: or \\ k_2 =  -  \frac{5}{2}  -  \frac{3 \sqrt{3} }{2}  \: i

Hope this helps you

8 0
3 years ago
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zepelin [54]

Answer:

its 3

Step-by-step explanation:

8 0
3 years ago
Similar triangles and polygons worksheet<br> help with number 1
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3 years ago
Solve 2(x + 1) = 2x + 2.<br> A All real numbers<br> B No solution<br> C 0<br> D 1
FrozenT [24]
2(x+1)=2x+2  if you factor the right side ..

2(x+1)=2(x+1)

This is always true for any value of x so there are infinitely many solutions...

x= all real numbers.
6 0
3 years ago
Read 2 more answers
Some help figuring out the answer?? Also explain a little how you got there
ipn [44]

9514 1404 393

Answer:

  x = 10·cos(θ) -4·cot(θ)

Step-by-step explanation:

Apparently, we are to assume that the horizontal lines are parallel to each other.

The relevant trig relations are ...

  Sin = Opposite/Hypotenuse

  Cos = Adjacent/Hypotenuse

If the junction point in the middle of AB is labeled X, then we have ...

  sin(θ) = 4/BX   ⇒   BX = 4/sin(θ)

  cos(θ) = x/XA   ⇒   XA = x/cos(θ)

Then ...

  BX +XA = AB = 10

Substituting for BX and XA using the above relations, we get

  4/sin(θ) +x/cos(θ) = 10

Solving for x gives ...

  x = (10 -4/sin(θ))·cos(θ)

  x = 10·cos(θ) -4·cot(θ) . . . . . simplify

_____

We used the identity ...

   cot(θ) = cos(θ)/sin(θ)

6 0
3 years ago
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