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2 years ago

What are the zeros(x+12)^2=0

Mathematics

2 answers:

dybincka [34]2 years ago

6 0

(x+12)^2 = 0

x + 12 = 0

x = - 12

wariber [46]2 years ago

4 0

(x+12)^2 = 0

x + 12 = 0

x = - 12

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Please help asap! Due tonight!

Contact [7]

Answer:

A, $\sqrt{17}$

Step-by-step explanation:

a^2 + b^2 = c^2

8^2 + b^2 = 9^2

64 + b^2 = 81

b^2 = 17

b = $\sqrt{17}$

4 0

2 years ago

Read 2 more answers

Points R, S, and T are collinear, and S is between R and T, in addition, RS = 2x-4, ST = 3x+2, and RT = 13. Find the value of RS

Alex17521 [72]

Answer:

RS = 2

Step-by-step explanation:

Points R, S, and T are collinear, and S is between R and T, in addition, RS = 2x-4, ST = 3x+2, and RT = 13. Find the value of RS

Hence:

RS + ST = RT

2x - 4 + 3x + 2 = 13

2x + 3x - 4 + 2 = 13

5x - 2 = 13

5x = 13 + 2

5x = 15

x = 15/5

x = 3

To find the value of RS

RS = 2x-4

RS = 2 × 3 - 4

RS = 6 - 4

RS = 2

5 0

2 years ago

sylvia rented a boat for $16.50 per hour. Her total rental fee was 49.50. For how many hours did Sylvia rent the boat

Finger [1]

49.50/16.50=3 hours is your answer

3 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

2 years ago

NEED HELP ASAP, I'LL GIVE BRAINLIEST. (EXPLANATION NEEDED) ALGEBRA

katrin [286]

$31x^2 +14y -15x +3\\\\=x(31x-15)+14y+3$

4 0

1 year ago