All categories

3 years ago

Not sure if I got it right...I know it is one of the first 2.

Mathematics

2 answers:

Ahat [919]3 years ago

4 0

Answer:

D. $\frac{-6x^{2}-x-7 }{x^{2}+9}$

Step-by-step explanation:

since the $x^{2}+9=x^{2} +9$ you just have to add $\frac{-6x^{2}+x-3-2x-4 }{x^{2}+9 }$ so the answer will be: $\frac{-6x^{2}-x-7 }{x^{2}+9 }$

e-lub [12.9K]3 years ago

3 0

Answer:

Step-by-step explanation:

-6x^2 + x -3 + 2x + 4

________________________

x^2 + 9

-6x^2 + 3x + 1

__________

x^2 + 9

the solution is the first one

You might be interested in

Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p

posledela

Answer:

P=0.0000037

P=0.00037%

Step-by-step explanation:

Probability

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

$\displaystyle \frac{4}{52}=\frac{1}{13}$

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

$\displaystyle \frac{3}{51}=\frac{1}{17}$

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

$\displaystyle \frac{2}{50}=\frac{1}{25}$

Finally, the probability of getting the remaining ace out of the 49 cards is:

$\displaystyle \frac{1}{49}$

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

$\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}$

$\displaystyle P= \frac{1}{270,725}$

P=0.0000037

P=0.00037%

3 0

2 years ago

working together 9 friends pick up 23 2/5 bags of apples and an orchard they divide the bags of apples equally between them how

Citrus2011 [14]

Simply divide 23 and 2/5 bags of apples between 9 friends:

23 and 2/5 bags ÷ 9 friends= 117/5 bags ÷ 9 friends
= 13/5 bags
= 2 and 3/5 bags

Therefore, each friend receives 2 and 3/5 bags of apples.

Hope this helps!

5 0

2 years ago

Two certificates of deposit pay interest that differ by 3%. Money invested for one year in the first CD earns $240 interest. The

Soloha48 [4]

a = interest rate of first CD

b = interest rate of second CD

and again, let's say the principal invested in each is $X.

$\bf a-b=3\qquad \implies \qquad \boxed{b}=3+a~\hfill \begin{cases} \left( \frac{a}{100} \right)X=240\\\\ \left( \frac{b}{100} \right)X=360 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \left( \cfrac{a}{100} \right)X=240\implies X=\cfrac{240}{~~\frac{a}{100}~~}\implies X=\cfrac{24000}{a} \\\\\\ \left( \cfrac{b}{100} \right)X=360\implies X=\cfrac{360}{~~\frac{b}{100}~~}\implies X=\cfrac{36000}{b} \\\\[-0.35em] ~\dotfill\\\\$

$\bf X=X\qquad thus\qquad \implies \cfrac{24000}{a}=\cfrac{36000}{b}\implies \cfrac{24000}{a}=\cfrac{36000}{\boxed{3+a}} \\\\\\ (3+a)24000=36000a\implies \cfrac{3+a}{a}=\cfrac{36000}{24000}\implies \cfrac{3-a}{a}=\cfrac{3}{2} \\\\\\ 6-2a=3a\implies 6=5a\implies \cfrac{6}{5}=a\implies 1\frac{1}{5}=a\implies \blacktriangleright 1.2 = x\blacktriangleleft$

$\bf \stackrel{\textit{since we know that}}{b=3+a}\implies b=3+\cfrac{6}{5}\implies b=\cfrac{21}{5}\implies b=4\frac{1}{5}\implies \blacktriangleright b=4.2 \blacktriangleleft$