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lina2011 [118]
3 years ago
5

K^2+8k-9=0 can you help me solve with showing work

Mathematics
2 answers:
givi [52]3 years ago
7 0
Using the PSN method:

P = -9
S = 8
So N = 9,-1

Using 9 and -1 in the original equation gives:

k^2+9k-k-9=0
\\k(k+9)-1(k+9)=0
\\(k-1)(k+9)

\\\\k=1,-9
erastova [34]3 years ago
5 0
K² + 8k - 9 = 0
k = <u>-8 +/- √(8² - 4(10)(-9))</u>
                   2(1)
k = <u>-8 +/- √(64 + 360)</u>
                   2
k = <u>-8 +/- √(424)
</u>               2<u>
</u>k = <u>-8 +/- 20.59126028
</u>                     2<u>
</u>k = <u>-8 + 20.59126028</u>    k = <u>-8 - 20.59126028
</u>                    2<u />                                    2<u>
</u>k = <u>12.5912608</u>             k = <u>-20.5912608
</u>              2    <u /><u />                                 2<u>
</u>k = 6.2956304<u />                k = <u />-10.2956304<u>
</u>
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First, you have to combine all the like terms such as both the 2x’s and 16 and 12 which will give us 28+4x=396. Then you subtract 396 from 28 which is 368. You want the variable x to be alone so you divide 368 by 4 which gives you 92.
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Does the following infinite series converge or diverge? 1/3+2/9+4/27+8/81
Norma-Jean [14]

Answer:

converge

Step-by-step explanation:

The ratio that you multiply by to get the next term is 2/3, which is less than 1, so the geometric series converges by the Geometric Series Test.

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1. Consider the right triangle ABC given below.
lbvjy [14]
#1) 
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
m∠A = 83°
a = 15.11

Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse.  The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25

Multiply both sides by 25:
25*sin 25 = (b/25)*25
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10.57 = b

B) The first way we can find a is using the Pythagorean theorem.  In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625

Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751

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√a² = √513.2751
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The second way is using the cosine ratio, adjacent/hypotenuse.  Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25

Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a

The third way is using the other angle.  First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°

Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65

Multiply both sides by 25:
(a/25)*25 = 25*sin 65
a = 25*sin 65
a = 22.66

#2)
A) Let side a be the one across from the 15° angle.  This would make the 15° angle ∠A.  We will define b as the side marked 4 and c as the side marked 3.  We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
a² = 25-24cos 15
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Take the square root of both sides:
√a² = √1.82
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Use the law of sines to find m∠B:
sin A/a = sin B/b
sin 15/1.35 = sin B/4

Cross multiply:
4*sin 15 = 1.35*sin B

Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
(4*sin 15)/1.35 = sin B

Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
50.07 = B

Subtract both known angles from 180 to find m∠C:
180-(15+50.07) = 180-65.07 = 114.93°

B)  Use the law of sines to find side b:
sin C/c = sin B/b
sin 52/12 = sin 45/b

Cross multiply:
b*sin 52 = 12*sin 45

Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
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Find m∠A by subtracting both known angles from 180:
180-(52+45) = 180-97 = 83°

Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a

Cross multiply:
a*sin 52 = 12*sin 83

Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
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\boxed{\sf sin\Theta=\dfrac{P}{H}}

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