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3 years ago

14

What is needed to make a rotation

2 answers:

Elden [556K]3 years ago

5 0

Answer:

the points of the graph

Step-by-step explanation:

there is a rule for every rotation for example a 90 degree rotation's rule is A(x,y) = A'(-y,x) to use them you need to have the points

marishachu [46]3 years ago

5 0

The points of the graph

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Y=-1/3x^2-4x-5 in vertex form

grigory [225]

Answer:

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

Step-by-step explanation:

Given function:

$Y=-\frac{1}{3}x^2-4x-5$

We need to find the vertex form which is.,

$y=a(x-h)^2+k$

where $(h,k)$ represents the co-ordinates of vertex.

We apply completing square method to do so.

We have

$Y=-\frac{1}{3}x^2-4x-5$

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

$-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5$

$-3Y=x^2+12x+15$

Isolating $x^2$ and $x$ terms on one side.

Subtracting both sides by 15.

$-3Y-15=x^2+12x-15-15$

$-3Y-15=x^2+12x$

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of $x$ term, square it and add it both sides side.

square of half of the co-efficient of $x$ term = $(\frac{1}{2}\times 12)^2=(6)^2=36$

Adding 36 to both sides.

$-3Y-15+36=x^2+12x+36$

$-3Y+21=x^2+12x+36$

Since $x^2+12x+36$ is a perfect square of $(x+6)$ , so, we can write as:

$-3Y+21=(x+6)^2$

Subtracting 21 to both sides:

$-3Y+21-21=(x+6)^2-21$

$-3Y=(x+6)^2-21$

Dividing both sides by -3.

$\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}$

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

8 0

3 years ago

I need help with no solution

drek231 [11]

use photo math For that Problem It would work Mate.

4 0

3 years ago

Read 2 more answers

PLASE ANSWER THIS QUESTION I GIVE YOU 18 POINTS

sukhopar [10]

Answer:

(a) 144

(b) 117

(c) 360

(d) 588

(e) 5472

Step-by-step explanation:

To find the LCM of two numbers, first find the prime factorization of each number. Then the LCM is the product of common and not common factors with the larger exponent.

2.

(a)

$72 = 2^3 \times 3^2$

$144 = 2^4 \times 3^2$

$LCM = 2^4 \times 3^2 = 8 \times 9 = 144$

(b)

$39 = 3 \times 13$

$117 = 3^2 \times 13$

$LCM = 3^2 \times 13 = 117$

(c)

$72 = 2^3 \times 3^2$

$90 = 2 \times 3^2 \times 5$

$LCM = 2^3 \times 3^2 \times 5 = 360$

(d)

$84 = 2^2 \times 3 \times 7$

$147 = 3 \times 7^2$

$LCM = 2^2 \times 3 \times 7^2 = 588$

(e)

$152 = 2^3 \times 19$

$288 = 2^5 \times 3^2$

$LCM = 2^5 \times 3^2 \times 19 = 5472$

6 0

3 years ago

Please help will give brainliest

butalik [34]

Answer:

360 m is the answer........

6 0

3 years ago

Read 2 more answers

PLEASE HELP MIDDLE SCHOOL MATH

snow_tiger [21]

Answer:

x>=5

Step-by-step explanation:

If x is less than five, then there will be a negative in the square root, which will yield x an imaginary number.

3 0

3 years ago