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3 years ago

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ABCD is a rectangle, where A (1, 2), B (6, 0), C (10,10) and D (x, y) is unknown. Find the coordinates of the fourth vertex Poin

t D. Verify that ABCD is a rectangle providing evidence related to the sides and angles. 2. Find the perimeter and area

1 answer:

Korolek [52]3 years ago

6 0

Answer:

So find the what?

Step-by-step explanation:

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Solve <br> 1)3x-7=35?<br> 2)22=7-3a<br> 3)-d+7=3<br> 4)2x+23=49<br> 5)-c+2=5

astra-53 [7]

$1)\ 3x-7=35\ \ \ |add\ 7\ to\ both\ sides\\\\3x=42\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=14}\\-------------------------\\2)\ 22=7-3a\\\\7-3a=22\ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-3a=15\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\\\\\boxed{a=-5}\\--------------------------$

$3)\ -d+7=3\ \ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-d=-4\\\\\boxed{d=4}\\---------------------------\\\\4)\ 2x+23=49\ \ \ \ \ \ |subtract\ 23\ from\ both\ sides\\\\2x=26\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=13}\\---------------------------\\5)\ -c+2=5\ \ \ \ \ \ |subtract\ 2\ from\ both\ sides\\\\-c=3\\\\\boxed{c=-3}$

7 0

3 years ago

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Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

3 years ago

Consider the quadratic function f(x)= 2(x + 4)² − 6. What is the vertex of the parabola? *

Masja [62]

Answer:

vertex is (-4,-6)

Step-by-step explanation:

put x+4=0

x=-4

when x=-4

f(x)=-6

4 0

4 years ago

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80% of what number is 5

seropon [69]

Answer:

1

Step-by-step explanation:

80% = 0.80

0.80 x 5 = 4

5 - 4 = 1

7 0

4 years ago

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Why are these triangles similar?

nataly862011 [7]

Answer SAS

Step-by-step explanation:

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3 years ago