What is the area of the triangle to the nearest tenth?

ABC has coordinates of A(-8,-8) B(4,-2) C(2,2). Find the coordinates of its image after the dilation centered at the origin with

garri49 [273]

Answer:

A' ( -12 , -12 )

B' ( 6 , -3 )

C' ( 3 , 3 )

Step-by-step explanation:

To find the coordinates of a point after a dilation simply multiply the x and y values of the points by the scale factor

Points: A(-8,-8) B(4,-2) C(2,2)

Scale factor: 1.5

Coordinates after the dilation

A' = (-8,-8) --------> (-8 * 1.5 , -8 * 1.5 ) ------------> (-12 , -12)

B' = (4,-2) ---------> (4 * 1.5 , -2 * 1.5) -----------> (6 , -3 )

C' = (2 , 2) ----------> (2 * 1.5 , 2 * 1.5) -----------> (3, 3)

So inclusion the coordinates of ABC after a dilation centered at the origin with a scale factor of 1.5 are A' ( -12 , -12 ) B' ( 6 , -3 ) C' ( 3 , 3 )

4 0

2 years ago

Joe is building a post for his mailbox. To find the correct dimensions, he needs to expand this expression: (x-3) (x-9) (x-4) Se

kolbaska11 [484]

Answer:

$(x-3)(x-9)(x-4)=x^3-16x^2+75x-108$

Step-by-step explanation:

Given:

The expression to expand is given as:

$(x-3)(x-9)(x-4)$

Let us expand the first two binomials of the given expression using FOIL method.

The FOIL method states that:

$(a + b)(c + d) = ac + ad + bc + bd$

$(x-3)(x-9)=(x\times x)+(x\times -9)+(-3\times x)+(-3\times -9)\\\\(x-3)(x-9)=x^2-9x-3x+27=x^2-12x+27$

Now, let us multiply the result with the remaining binomial. Multiplying each term of the trinomial with each term of the binomial, we get:

$(x^2-12x+27)(x-4)\\\\= (x^2\times x)+ (x^2\times -4)+(-12x\times x)+(-12x\times -4)+(27\times x)+(27\times -4)\\\\=x^3-4x^2-12x^2+48x+27x-108\\\\=x^3-16x^2+75x-108...........(\because-12x^2-4x^2=-16x^2,48x+27x=75x)$

Therefore, the equivalent expression after expanding is given as:

$(x-3)(x-9)(x-4)=x^3-16x^2+75x-108$

3 0

2 years ago

Need asap please!!!!

kirill [66]

Answer:

use ur brain

Step-by-step explanation:

ask ur teacher if the be bad teacher google it

7 0

3 years ago

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

8 0

3 years ago

The gross weight of a package and its contents is 11 pounds 5 ounces. If the packaging weighs 1 pound 15 ounces, what's the net

dsp73

Answer A
For ease convert both into ounces.

Gross = (11x16) + 5 = 181 ounces
Packaging = (1x16) + 15 = 31 oz.

Now subtract Packaging from Gross

181 - 31 = 150 oz.

Divide by 16 to get pounds

150/16 = 9.375 - hence 9 lb.

and multiply 16 by .375 for ounces

0.375 x 16 = 6 oz.

Thus A : 9lb 6 oz

5 0

3 years ago

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