Please Asap. Find the value of s.

Solve on the interval (0,pi) 1+cosθ=((sqrt3)+2)/2

Oliga [24]

$\bf 1+cos(\theta )=\cfrac{\sqrt{3}+2}{2}\implies cos(\theta )=\cfrac{\sqrt{3}+2}{2}-1
\\\\\\
cos(\theta )=\cfrac{\sqrt{3}+2}{2}-\cfrac{2}{2}\implies cos(\theta )=\cfrac{\sqrt{3}\underline{+2-2}}{2}\implies cos(\theta )=\pm\cfrac{\sqrt{3}}{2}
\\\\\\
cos^{-1}[cos(\theta )]=cos^{-1}\left( \pm\cfrac{\sqrt{3}}{2} \right)\implies \measuredangle \theta =
\begin{cases}
\frac{\pi }{6}\\\\
\frac{5\pi }{6}\\\\
\frac{7\pi }{6}\\\\
\frac{11\pi }{6}
\end{cases}$

8 0

3 years ago

My brain is lagging:<

Lubov Fominskaja [6]

Distance between two points $(-3, 4)$ and $(5, -2)$ is $10$ units.

Mid-point of the line segment joining $(-1, -5)$ and $(7, 1)$ is $(3, -2)$ .

Degree of the given polynomial is $7$ .

6) Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So, distance between two points $(-3, 4)$ and $(5, -2)$ is:

$=\sqrt{(5-(-3))^2+(-2-4)^2}$

$=\sqrt{(8)^2+(-6)^2}$

$=\sqrt{64+36}$

$=\sqrt{100}$

$=10$

So, option D is correct.

7) Mid-point of the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is:

$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Mid-point of the line segment joining $(-1, -5)$ and $(7, 1)$ is:

$(\frac{-1+7}{2}, \frac{-5+1}{2})$

$= (\frac{6}{2}, \frac{-4}{2})$

$= (3, -2)$

So, option A is correct.

8) Degree of a polynomial is the highest power of the variable in the polynomial.

$f(x)=3x^2-7x^4+9x^7+2x^3$

Here, the highest power of the variable is $7$ .

So, option B is correct.

Learn more about distance formula, mid point formula, and degree of polynomial.

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8 0

2 years ago

Please help me I need help please

MakcuM [25]

Answer:

p = 4

Step-by-step explanation:

Given equation:

$x^2+(p-3)y^2-4x+6y-16=0$

Standard equation of a circle:

$(x-a)^2+(y-b)^2=r^2$

(where $(a,b)$ is the centre of the circle, and $r$ is the radius)

If you expand this equation, you will see that the coefficient of $y^2$ is always one.

Therefore, $p-3=1$

$\implies p=1+3=4$

Additional information

To rewrite the given equation in the standard form:

$\implies x^2+y^2-4x+6y-16=0$

$\implies x^2-4x+y^2+6y=16$

$\implies (x-2)^2-4+(y+3)^2-9=16$

$\implies (x-2)^2+(y+3)^2=16+4+9$

$\implies (x-2)^2+(y+3)^2=29$

So this is a circle with centre (2, -3) and radius √29

3 0

2 years ago

The length of a rectangle is 6 yd longer than its width. if the perimeter of the rectangle is 44 yd , find its area.

NNADVOKAT [17]

Let's call the length L and the width w.

perimeter = 2L + 2w
44 = 2L + 2w

the length is 6 yards longer than the width. so L = w + 6.

plug this into the equation. 44 = 2(w+6) + 2w
44 = 2w + 12 + 2w
44 = 4w + 12
32 = 4w
w = 8
L = w + 6
L = 8 + 6
L = 14

area = Lw
area = 8(14)
area = 112 yd²

3 0

3 years ago

The total square footage of mini storage units being rented in a city is graphed to forecast future demand. The scatter plot has

Monica [59]

Answer:

B 160,000 sq ft

Step-by-step explanation:

2018 is 13 years after 2005.

Put x=13 in the equation and evaluate:

... y = 0.538·13 +9.02 = 6.994 +9.02 = 16.014

This is the number of 10,000 sq ft, so the required area is ...

... 16 times 10,000 sq ft = 160,000 sq ft.

5 0

3 years ago

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