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3 years ago

If 2n = 54/9 then n = ______

Mathematics

1 answer:

Fantom [35]3 years ago

8 0

Answer:

1). 3

2). 6

3). 9

Step-by-step explanation:

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Can someone please help

choli [55]

The answer is C, 2 over 11.

There are 11 letters in Mississippi, and there is 2 p's, there fore, he has a 2 out of 11 chance to get a P.

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3 years ago

The conditional relative frequency table was generated using data that compares the number of boys and girls who pack their lunc

pochemuha

Answer:

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2 years ago

Find the area of the figure.

givi [52]

these are just two semi-circles, one with a diameter of 22 and the other with a diameter of 30 - 22, or 8, and therefore their radius are 11 and 4 respectively.

$\bf \textit{area of a semi-circle}\\\\ A=\cfrac{\pi r^2}{2}~~,~~ \stackrel{r=11}{A=\cfrac{\pi 11^2}{2}}\implies A=\cfrac{121\pi }{2}~~,~~ \stackrel{r=4}{A=\cfrac{\pi 4^2}{2}}\implies A=8\pi \\\\ -------------------------------\\\\ \stackrel{\textit{total area of the figure}}{\cfrac{121\pi }{2}+8\pi }\implies \cfrac{121\pi +16\pi }{2}\implies \cfrac{137\pi }{2}$

4 0

3 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

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1 year ago

Determinar el área de un hexágono que esta inscrito en una circunferencia de 4 unidades de radio

Dmitrij [34]

Answer:

espero y te sirva hay esta todo lo que necesitas

6 0

2 years ago