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What does a² = 58 - 42 cos 35   ?

$$\begin{array}{ccccl}
a^2 &=& b^2+c^2 & -2bc & *\cos{ ( \alpha ) }\\
a^2 &=& 58 &-42 & *\cos{ ( 35\ensurement{^{\circ}} ) }
\end{array}$$

$$\\ \small{\text{
(1)
}}
\ b^2+c^2 = 58
\\
\small{\text{
(2)
}}
\ 2bc = 42 \small{\text{ or }} c = \frac{21}{b}$$

$$\small{\text{
(2) in (1):
}}
\ b^2 + \frac{(21^2)} {b^2} = 58 \small{\text{ or }} b^4-58b^2+21^2=0 \\ \small{\text{ if }} x=b^2 \small{\text{ then }} x^2-58x+21^2=0 \\
x_{1,2} = \frac{58\pm40}{2} \small{\text{ so }} x_1=49 \small{\text{ and }} x_2 = 9 \small{\text{ because }} b=+\sqrt{x} \\
\small{\text{ so }} b_1 = 7 \small{\text{ or }} b_2 = 3 \\
c=\frac{21}{b} \small{\text{ so }} c_1 = 3 \small{\text{ or }} c_2 = 7$$

Hi DragonSlayer554,

compare

Solve on the interval [0,2pi): 4cscx+6=-2

$$\\ 4 \csc{(x)}+6=-2 \quad | \quad -6 \\
4 \csc{(x)}=-8 \quad | \quad :2 \\
\csc{(x)}=-2 \quad | \quad \csc{(x)} = \frac{1}{\sin{(x)}} \\
\frac{1}{ \sin{(x)} } = -2 \\
\sin{(x)} = -\frac{1}{2}$$

$$x_1 = \sin^{-1}{(-\frac{1}{2} )} \\
x_1 = - \sin^{-1}{(\frac{1}{2} )} \\
x_1 = -30 \ensurement{^{\circ}} + 360 \ensurement{^{\circ}} = 330 \ensurement{^{\circ}}$$

$$x_2 = 180 \ensurement{^{\circ}} - (-30 \ensurement{^{\circ}} ) = 210 \ensurement{^{\circ}}$$

Find the cube root of 8*7*9*49*3

$$\sqrt[3]{8*7*9*49*3} = \sqrt[3]{2^33^37^3} = 2*3*7 = 42$$

3/4y-2/3x=12. Find the x and y intercept

$$\begin{array}{rcl}
\dfrac{3}{4}y-\dfrac{2}{3}x &=& 12 \quad | \quad :12 \\ \\
\dfrac{1}{16}y-\dfrac{1}{18}x &=& 1 \\ \\
\dfrac{y}{16}+\dfrac{x}{-18} &=& 1
\end{array}$$

$$\small{ \text{
y-intercept is 16 and x-intercept is -18
}
}$$

25 daL to kiloliters

$$25\ daL
*
\left(
\dfrac{10^1} {1\ da}
\right)
*
\left(
\dfrac{1\ k} {10^3}
\right)
=25
*
\left(
\dfrac{10} {10^3}
\right)
kL
=
\dfrac{25} {100}
kL
=0.25\ kL$$

Is this correct?

find an equation of the line. write the equation using function notation. through (7,-3) and perpindicular to 9y=x-18.

$$9y=x-18 \quad \small{\text{ or }} \quad
y = \frac{1}{9}x - 2 \quad \small{\text{ the slope = }} m = \frac{1}{9}$$

$$\small{\text{the slope of the perpindicular line = }}
m_{\perp} = -\frac{1}{m} = -\dfrac{1}{ \frac{1}{9} } = -9$$

$$\small{\text{the function of the perpindicular line = }}
f(x) = y = m_{\perp}(x-x_p)+y_p$$

$$\\\small{\text{with point }} x_p = -3 \small{\text{ and }} y_p = 7 }} \small{\text{, }}
f(x) = y = -9 [x-(-3)]+7 \\
y= -9(x+3)+7 \\
y= -9x-27+7 \\
\boxed{y= -9x -20}$$

x(1-3x) + 3x-1

=-x(3x-1) +(3x-1)

=(3x-1)-x(3x-1)

=(3x-1)(1-x)

((1/2)x-3)^4

$$=\left(\frac{x}{2}-3\right)^4=\left(\frac{x-6}{2}\right)^4=\frac{1}{16}(x-6)^4$$

$$=\frac{1}{16}
( \
\textcolor[rgb]{1,0,0}{1} *x^{\textcolor[rgb]{0,1,0}{4}}*6^{\textcolor[rgb]{0,0,1}{0}}
-\textcolor[rgb]{1,0,0}{4} *x^{\textcolor[rgb]{0,1,0}{3}}*6^{\textcolor[rgb]{0,0,1}{1}}
+\textcolor[rgb]{1,0,0}{6} *x^{\textcolor[rgb]{0,1,0}{2}}*6^{\textcolor[rgb]{0,0,1}{2}}
-\textcolor[rgb]{1,0,0}{4} *x^{\textcolor[rgb]{0,1,0}{1}}*6^{\textcolor[rgb]{0,0,1}{3}}
+\textcolor[rgb]{1,0,0}{1} *x^{\textcolor[rgb]{0,1,0}{0}}*6^{\textcolor[rgb]{0,0,1}{4}}
\ )$$    $$\left[ \small{\text{Pascal 4: 1,4,6,4,1}} \right]$$

$$=\frac{1}{16}
(\
x^4-4*x^3*6+6*x^2*6^2-4*x*6^3+6^4
\ )$$

$$=\frac{1}{16}
( \
x^4-24x^3+216x^2-864x+1296
\ )$$

$$=
\frac{1}{16}x^4
-\frac{3}{2} x^3
+\frac{27}{2}x^2
-54x
+81$$