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Hitman42 [59]
3 years ago
5

Hi there! i going to fail if you don´t answer :D :D

Mathematics
1 answer:
NARA [144]3 years ago
5 0

Answer:

it will be 1/2

Step-by-step explanation:

because x×y={(6,2), (6, 8),(9,2), (9, 8)}

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Convert the Quadratic Function below from Standard Form to Vertex Form.* y = x2 – 2x – 3 A. y = (x - 1)2 + 4
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Answer: the answer is B

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Frank left his house at 7 a.m. and drove to the airport at a speed of 50 mph. Lance left his house at 6 a.m. and drove to the sa
frosja888 [35]

Answer:

10 AM

Step-by-step explanation:

Frank drove at a speed of 50 mph, and traveled a distance of 150 miles.  This means he traveled for

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3 0
2 years ago
The perimeter of 2ft squared
Brums [2.3K]

Answer:

the peremiter will be 2 x one factor +2 the other factor. lets say the area of the rectangle is square feet. then the rectangle could be 8x12; so the perimeter would be 2x8+2×12 or 40 but the rectangle could also be 6 x 16 the peremiter would be 44

4 0
3 years ago
If the endpoints of the diameter of a circle are (8, −6) and (4, −2), what is the standard form equation of the circle?
Gennadij [26K]
We do the midpoint formula to find the center of the circle to get the left side of the equation.

Midpoint = [(X₁ + X₂) / 2 , (Y₁ + Y₂) / 2] 

Now plug in:

[(8 + 4) / 2 , (- 6 - 2) / 2]
(12 / 2 , - 8 / 2)
(6, - 4)

The center of the circle is (6, -4)

Now we plug it into the equation of a circle:

(x - h)² + (y - k)² = r² where (h, k) is the center of the circle and r is the radius. 

(x - 6)² + (y + 4)² = r² is the left side of the equation. This will eliminate options A and C

Now we do the distance formula using the center and an endpoint to get the radius. The formula for distance is:

√((X₂ - X₁)² + (Y₂ - Y₁)²)

We plug in using either of the endpoints. 

√((4 - 6)² + (- 2 - (- 4))²)
√((-2)² + 2²)
√(4 + 4)
√8

√8 is your radius

(√8)² = 8

Your correct answer is (x - 6)² + (y + 4)² = 8, Option D
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cint%5C%20%28x%5E3-6x%5E2%2B9x%2B3%29%283x%5E2-12x%2B9%29%20dx" id="TexFormula1" title="\int
AlexFokin [52]

If the integral is simply

\displaystyle\int(x^3-6x^2+9x+3)(3x^2-12x+9)\,\mathrm dx

then notice that

\mathrm d(x^3-6x^2+9x+3)=(3x^2-12x+9)\,\mathrm dx

which means you can compute the integral easily with a substitution

u=x^3-6x^2+9x+3\implies\mathrm du=(3x^2-12x+9)\,\mathrm dx

Under this transformation, the integral is

\displaystyle\int u\,\mathrm du=\frac{u^2}2+C=\boxed{\frac{(x^3-6x^2+9x+3)^2}2+C}

On the other hand, in case you're missing a symbol and the integral is actually

\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx

then first carry out the division:

\dfrac{x^3-6x^2+9x+3}{3x^2-12x+9}=\dfrac x3-\dfrac23-\dfrac{2x-9}{3x^2-12x+9}

Now, 3x^2-12x+9=3(x-3)(x-1), so to integrate the remainder term you can decompose it into partial fractions:

-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac a{x-3}+\dfrac b{x-1}

9-2x=a(x-1)+b(x-3)

x=1\implies7=-2b\implies b=-\dfrac72

x=3\implies3=2a\implies a=\dfrac32

\implies-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac 3{2(x-3)}-\dfrac 7{2(x-1)}

Then the integral would be

\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx=\boxed{\frac{x^2}6-\frac{2x}3+\frac32\ln|x-3|-\frac72\ln|x-1|+C}

which can be rewritten in several ways, such as

\dfrac{x^2-4x}6+\dfrac12ln\left|\dfrac{(x-3)^3}{(x-1)^7}\right|+C

6 0
3 years ago
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