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3 years ago

Here all me points i have 0 now

Mathematics

1 answer:

marin [14]3 years ago

4 0

Answer:

thanks

Step-by-step explanation:

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What is 5 + 10 pls pls pls pls pls help

AlekseyPX

Answer:

Step-by-step explanation:

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3 years ago

-19.8 > 3.6y<br><br> Can someone help solve this inequality please?

Step2247 [10]

-19.8 > 3.6y

-19.8/3.6 > 3.6y/3.6

-5.5 > y

4 0

3 years ago

Aisha has 9 pizzas she is serving each person at a party 3/4 of a pizza.

Nikitich [7]

Answer:

Step-by-step explanation:

9/0.75=12.

7 0

3 years ago

☆pls help no one has helped me and I'm so confused!

motikmotik

Answer:

Step-by-step explanation:

a). $\frac{5\pm\sqrt{-4}}{3}$ = $\frac{5}{3}\pm \frac{\sqrt{-4}}{3}$

= $\frac{5}{3}\pm\frac{2\sqrt{(-1)} }{3}$

= $\frac{5}{3}\pm\frac{2i}{3}$ [Since, i = $\sqrt{(-1)}$ ]

b). $\frac{10\pm\sqrt{-16}}{2}$ = $\frac{10}{2}\pm \frac{\sqrt{-16}}{2}$

= $5\pm \frac{4\sqrt{-1}}{2}$

= 5 ± 2i [Since, i = $\sqrt{(-1)}$ ]

c). $\frac{-3\pm \sqrt{-144}}{6}$ = $-\frac{3}{6}\pm \frac{\sqrt{-144}}{6}$

= $-\frac{3}{6}\pm \frac{12\sqrt{-1}}{6}$

= $-\frac{1}{2}\pm 2\sqrt{-1}$

= $-\frac{1}{2}\pm 2i$ [Since, i = $\sqrt{(-1)}$ ]

4 0

3 years ago

Rewrite in Polar Form....<br> x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

3 0

3 years ago