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2 years ago

x^2=6x+1 rewrite the equation by completing the square your equation should look like (x+a)^2=b or (x-c)^2=d

Mathematics

1 answer:

ryzh [129]2 years ago

8 0

Answer:

(x - 3)² = 10

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Completing the Square

Step-by-step explanation:

Step 1: Define

x² = 6x + 1

Step 2: Rewrite

Subtract 6x on both sides: x² - 6x = 1
Complete the Square: x² - 6x + 9 = 1 + 9
Factor: (x - 3)² = 10

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Answer: 11. Is A

Step-by-step explanation:

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2 years ago

Which expression is equivalent to ( 4 1/3)^3 ?

shutvik [7]

(13/3)^3 or (13/3)•(13/3)•(13/3)

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2 years ago

9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

SSSSS [86.1K]

Answer:

Part 4) $r=84\ units$

Part 9) $sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) $sin(\theta)=-\frac{9\sqrt{202}}{202}$

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

$C=2\pi r$

using proportion

$\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}$

simplify

$\frac{r}{180^o}=\frac{56}{120^o}$

solve for r

$r=\frac{56}{120^o}(180^o)$

$r=84\ units$

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

$cos^2(\theta)+sin^2(\theta)=1$

we have

$cos(\theta)=-\frac{2}{3}$

substitute the given value

$(-\frac{2}{3})^2+sin^2(\theta)=1$

$\frac{4}{9}+sin^2(\theta)=1$

$sin^2(\theta)=1-\frac{4}{9}$

$sin^2(\theta)=\frac{5}{9}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{5}}{3}$

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

$sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?

see the attached figure to better understand the problem

In the right triangle ABC of the figure

$sin(\theta)=\frac{BC}{AC}$

Find the length side AC applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$AC^2=11^2+9^2$

$AC^2=202$

$AC=\sqrt{202}\ units$

$sin(\theta)=\frac{9}{\sqrt{202}}$

simplify

$sin(\theta)=\frac{9\sqrt{202}}{202}$

Remember that

The point (11,-9) lies in Quadrant IV

then

The value of sin(∅) is negative

therefore

$sin(\theta)=-\frac{9\sqrt{202}}{202}$

5 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

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Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$ .

Take the normal vector to $S$ to be

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$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$

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2 years ago

PLS ANSWER ASAPP!! THX

Anastasy [175]

TanA = 6/8
or 3/4 when simplified

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2 years ago