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3 years ago

Alicia borrowed $15,000 to buy a car. She borrowed the money at 8% for 6 years.

Mathematics

1 answer:

PtichkaEL [24]3 years ago

4 0

Answer:

the answer is $7200 if that is correct

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What is the probability of tossing two heads

vodka [1.7K]

Hello!

Each time, there will be a 1/2 chance of tossing a head. Therefore, we multiply our two probabilities.

1/2(1/2)=1/4

Therefore, we have a 1/4 or 25% chance of tossing two heads in a row.

I hope this helps!

3 0

3 years ago

Read 2 more answers

A cube labeled 1 to 6, is tossed 6 times

slava [35]

Answer:

1/6

Step-by-step explanation:

If we're assuming that this is a six-sided die, then we can find that the results can be 1,2,3,4,5,6. Only 1 of those 6 numbers are 5, so the probability of the cube landing on 5 would be 1/6.

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3 years ago

If each cube has edges .5 inches long, what is the volume of the prism outlined in blue?

likoan [24]

Answer:

B) 7.5 in3

Step-by-step explanation:

8 0

3 years ago

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7. Find the measure of one exterior angle of<br><br> the regular polygon.<br><br> Triangle

Phantasy [73]

Answer:120°

Step-by-step explanation:

sum of exterior angles=360

In a triangle we have 3 exterior angles

Size for each exterior angle=360/3

Size for each exterior angle=120°

8 0

3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago