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Nimfa-mama [501]

3 years ago

7

Rewrite the equation by completing the square. x^2+ 12x + 36 = 0

2 answers:

Maslowich3 years ago

5 0

Answer:

(x+6)^2 =0

Step-by-step explanation:

svetoff [14.1K]3 years ago

4 0

Answer:

(x+6)^2 =0

Step-by-step explanation:

x^2+ 12x + 36 = 0

Subtract 36 from each side

x^2+ 12x + 36-36 = 0-36

x^2 +12x = -36

Take the coefficient of x

12

Divide by 2

12/2 =6

Square it

6^2 =36

Add this to each side

x^2+ 12x + 36 = -36+36

(x+12/2)^2 = 0

(x+6)^2 =0

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The value of: [(27)^2/3 × (8)^2/3]/9^3/2 is<br>(a) 1/3<br>(b) 2/3<br>(c) 3/3<br>(d) 4/3

Sever21 [200]

Step-by-step explanation:

<h3>➤ <u>Given</u>:-</h3>

$\sf{\frac{(27 {)}^{ \frac{2}{3} } \times (8 {)}^{ \frac{2}{3} } }{(9 {)}^{ \frac{3}{2} } } } \\$

<h3>➤ <u>To </u><u>find</u><u>:</u><u>-</u></h3>

${\sf Simplifyed \:form = ?}\\$

<h3>➤ <u>Solution</u><u>:</u><u>-</u></h3>

$\textsf{We have,}\\$

$\sf{\frac{(27 {)}^{ \frac{2}{3} } \times (8 {)}^{ \frac{2}{3} } }{(9 {)}^{ \frac{3}{2} } } } \\$

$\sf \leadsto \: {\frac{( {3}^{3} {)}^{ \frac{2}{3} } \times ( { {2}^{3} )}^{ \frac{2}{3} } }{( {3}^{2} {)}^{ \frac{3}{2} } } } \\$

$\sf \leadsto \: \frac{ {3}^{ \cancel3 \times \frac{2}{ \cancel3} } \times {2}^{ \cancel3 \times \frac{2}{ \cancel3} } }{ {3}^{ \cancel2 \times \frac{3}{ \cancel2} } } \\$

$\sf \leadsto \: \frac{ {3}^{2} \times {2}^{2} }{(3 {)}^{3} } \\$

$\sf \leadsto \: \frac{ {2}^{2} }{ {3}^{(3 - 2)}} \\$

$\sf \leadsto \: \frac{ {2}^{2} } { {3}^{1} } \\$

$\sf \leadsto \: \frac{2 \times 2}{3} \\$

$\sf \leadsto \: \frac{4}{3} \\$

8 0

3 years ago

Derivative of sin(0.354x)

sasho [114]

Answer:

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8 0

3 years ago

Eliminate the parameter

Pani-rosa [81]

Option 1. $y = 3x^{2} +7$ is obtained by eliminating the parameter.

Step-by-step explanation:

Step 1:

We have that $x = \sqrt{t}$ and $y = 3t+7.$

Take $x = \sqrt{t}$ as equation 1.

Assume $y = 3t+7$ is equation 2.

We need to determine a value that is equivalent to t so it can be substituted in equation 2.

Step 2:

If we square equation 1, we get

$x^{2} = (\sqrt{t})^{2} , x^{2} =t .$

So we have a value that can replace t, so we substitute this is equation 2.

$y = 3t+7, x^{2} =t,$ so

$y = 3x^{2} +t$ which is the first option.

3 0

3 years ago

What is the value of 7 in:<br> a. 3.7<br> b. 12.07<br> c. 200.711

masya89 [10]

A. seven tents b. seven hundredths c. seven tents

6 0

3 years ago

Which function has a vertex at the origin?

Crank

Answer:

B) $f(x)=x(x-4)$

Step-by-step explanation:

To find the function which has vertex at the origin.

Solution:

For a function to have its vertex at origin the function must pass through point (0,0) which means $f(0)=0$ .

We will find $f(0)$ for each of the given functions and check if it has vertex at origin.

To find $f(0)$ we will plugin $x=0$ in the function.

A) $f(x)=x+4$

$f(0)=0+4=4$

Vertex is not at the origin.

B) $f(x)=x(x-4)$

$f(0)=0(0-4)=0$ [Any number times zero is = zero]

Vertex is at the origin.

C) $f(x)=(x-4)(x+4)$

$f(0)=(0-4)(0+4)=(-4)(4)=-16$

Vertex is not at the origin.

D) $f(x)=2$

$f(0)=2$

Vertex is not at the origin.

3 0

3 years ago