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3 years ago

How do you solve y=a(x-h)^2+k

Mathematics

1 answer:

Alexxx [7]3 years ago

3 0

Yes i think thats right the person that did it above

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Solve x2 − 12x + 5 = 0 using the completing-the-square method. x = six plus or minus the square root of five x = negative six pl

Andru [333]

we are given

$x^2-12x+5=0$

we have to solve it by completing square method

step-1: Move 5 on right side

$x^2-12x+5-5=0-5$

$x^2-12x=-5$

step-2: Break middle term

$x^2-2*6*x=-5$

step-3: Add 6^2 both sides

$x^2-2*6*x+(6)^2=-5+(6)^2$

$(x-6)^2=31$

step-3: Solve for x

$\sqrt{(x-6)^2} =\sqrt{31}$

we wil get two values

First value is

$x-6=\sqrt{31}$

add both sides 6

$x-6+6=6+\sqrt{31}$

$x=6+\sqrt{31}$

Second value is

$x-6=-\sqrt{31}$

add both sides 6

$x-6+6=6-\sqrt{31}$

$x=6-\sqrt{31}$

so, solutions are

$x=6+\sqrt{31}$

$x=6-\sqrt{31}$ .............Answer

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3 years ago

What is the probability of choosing a king in a deck of 52 cards?

nikklg [1K]

K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again.

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3 years ago

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What are complex numbers?

dimaraw [331]

Answer:

Pie is a complex number?

Step-by-step explanation:

3.14159265358979323846264338327950288419716939937510582097494459230781640628..............................

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3 years ago

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The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in fourth quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in 4th quadrant so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago

I need help on how to solve the problem and the correct answer :)

lakkis [162]

This is a system of linear equations. First, you can add the two equations together to eliminate the y so that you can solve for x:

-5x + -7x = 0 + -96

-12x = -96

x = 8

Use x to solve for y:

-5 * 8 + 8y = 0

Add 40 to both sides and divide by 8:

y = 5

So, x = 8 and y = 5.

3 0

4 years ago

How do you solve y=a(x-h)^2+k​

How do you solve y=a(x-h)^2+k