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2 years ago

8

AABC goes through a sequence of transformations to form ABC. The sequence of transformations involved is a followed by a

1 answer:

lbvjy [14]2 years ago

7 0

Answer:

Step-by-step explanation:

what he said was he right answer

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What property is this

Alexxandr [17]

I think this is the associative property of multiplication.

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4 years ago

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0.04 (decimal) fraction lowest term

RSB [31]

Answer:

Step-by-step explanation:

<u> 4 = 1 </u>

100 25

4 is the hundredths place, not hundreds because hundredths is a fraction of a whole. So its 4 over the value of its place which is 1 hundredths and 4 goes into itself 1 and 4 goes into 100, 25 times.

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4 years ago

If n(X)=200,x)=300 and n(XUY)=400

olga_2 [115]

Answer:

${ \boxed{ \bf{formular : { \tt{n(X{ \small{U}Y) = n(X) + n(Y) - n(X{ \small{n}Y)}}}}}}} \\ n(X{ \small{n}}Y) = (200 + 300) - 400 \\ = 500 - 400 \\ { \boxed{n(X{ \small{n}}Y) = 100}} \\ \\ { \underline{ \blue{ becker \: jnr}}}$

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3 years ago

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Find the solutions to x⁴-5x²-36=0 and the x-intercepts of the graph of y=x⁴-5x²-36.

disa [49]

Answer:

The solutions $x^4-5x^2-36=0$ are $x=3,\:x=-3,\:x=2i,\:x=-2i$ and the x-intercepts of $y=x^4-5x^2-36$ are $x=3,\:x=-3$

Step-by-step explanation:

Finding the solutions to $x^4-5x^2-36$ means finding the roots, a root is where the function is equal to zero.

The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero.

To find the roots you need to:

Rewrite the equation with $u=x^2$ and $u^2=x^4$

$u^2-5u-36=0$

Solve by factoring

$\mathrm{Break\:the\:expression\:into\:groups}\\u^2-5u-36=\left(u^2+4u\right)+\left(-9u-36\right)$

$\mathrm{Factor\:out\:}u\mathrm{\:from\:}u^2+4u=u\left(u+4\right)$

$\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9u-36=-9\left(u+4\right)$

$u^2-5u-36=u\left(u+4\right)-9\left(u+4\right)$

$\mathrm{Factor\:out\:common\:term\:}u+4\\\left(u+4\right)\left(u-9\right)$

$u^2-5u-36=\left(u+4\right)\left(u-9\right)=0$

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions to the quadratic equation are:

$\:u=-4,\:u=9$

Substitute back $u=x^2$ , solve for x

$x^4-5x^2-36=(u-9)(u+4)=(x^2-9)(x^2+4)$

Apply the difference of squares formula

$x^4-5x^2-36=(x^2-9)(x^2+4)=(x-3)(x+3)(x^2+4)$

$(x-3)(x+3)(x^2+4)=0$

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions are:

$\:x=3,\:x=-3,\:x=2i,\:x=-2i$

Because two of the solutions are complex roots the only x-intercepts are $x=3,\:x=-3$

3 0

3 years ago

At is 5/9 4/6 in simplest form?

S_A_V [24]

5/9 is already in it's simplest form

4/6 can be reduced to 2/3

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3 years ago