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2 years ago

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

2 answers:

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]2 years ago

5 0

K² + 8k - 9 = 0
k = -8 +/- √(8² - 4(10)(-9))
 2(1)
k = -8 +/- √(64 + 360)
 2
k = -8 +/- √(424)
 2
k = -8 +/- 20.59126028
 2
k = -8 + 20.59126028 k = -8 - 20.59126028
 2 2
k = 12.5912608 k = -20.5912608
 2 2
k = 6.2956304 k = -10.2956304

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A central angle in a circle with a radius of 3 cm intercepts an arc with a length of 6 cm what is the measure of the central ang

Fiesta28 [93]

Answer:

$114.59\°$

Step-by-step explanation:

we know that

The circumference of a circle is equal to

$C=2\pi r$

In this problem we have

$r=3\ cm$

substitute

$C=2\pi (3)=6 \pi\ cm$

Remember that

$360\°$ subtends the arc length of the complete circle

by proportion

Find the measure of the central angle for an arc length of $6\ cm$

$\frac{360}{6\pi } \frac{degrees}{cm} =\frac{x}{6} \frac{degrees}{cm} \\ \\x=6*360/(6\pi )\\\\ x=114.59\°$

7 0

2 years ago

4 thousands 7 hundreds = 47

motikmotik

If there is 4 thousands and 7 hundreds is 4,700

4 0

2 years ago

Solve the system of linear equations. 2x + 6y = -3 2x + 6y = 0 Ο Α (0, 2) infinitely many solutions C (9,7) no solution

galina1969 [7]

Answer:

Step-by-step explanation:

-3 = 2x+6y

2x+6y = 0

By transitivity, -3 = 0, a contradiction. No solution.

If you graph the lines, you will see they are parallel, therefore never intersect.

6 0

2 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

2 years ago

1) Q: A: If 5 + x = 12, and you add -5 to the left side of equation, what should you add to right side of equation

san4es73 [151]

You can "add" like you say or you could just call it subtracting. If you subtract 5 from 5 you still get zero. Then on the right side, you would do the same. You would subtract 5 from 12 getting x, which is 7.

If you added you would still get the same answers but it is easier to say just subtract. Anything you do to the left, do to the right.

Hope this helps :)

3 0

3 years ago

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