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3 years ago

20 points

Mathematics

1 answer:

irinina [24]3 years ago

4 0

Answer:

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

(b) $9a(2a+5)$

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

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

Separate the solutions

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

The equation has complex roots

Separate the real and imaginary parts

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

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Solve 2(x + 1) = 2x + 2. A All real numbers B No solution C 0 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.

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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)

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