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3 years ago

What is the axis of symmetry of a quadratic and how is it related to its vertex?

Mathematics

1 answer:

zheka24 [161]3 years ago

7 0

Answer:

It goes through the vertex

Step-by-step explanation:

The axis of symmetry is an imaginary line that runs vertically or horizontally through the vertex depending on which way the parabola or hyperbola is facing.

Brainliest please.

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7. Is a triangle with side lengths 9, 18, 7 a right triangle, an acute triangle, or

Reptile [31]

I think the answer is Obtuse

8 0

3 years ago

What is the volume of a sphere with a diameter of 9 in.?

a_sh-v [17]

Answer: 121.5 in³

Explanation:

$\textnormal{Volume }= \cfrac{4}{3} \pi (4.5)^3 = 121.5 \pi \ in^3$

6 0

3 years ago

Read 2 more answers

Dina invests $600 for 5 years at a rate of 2% per year compound interest.

Aleks [24]

Answer:

The value of this investment at the end of the 5 years is of $662.5.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Dina invests $600 for 5 years at a rate of 2% per year compound interest.

This means that $P = 600, t = 5, r = 0.02, n = 1$ . Thus

$A(t) = P(1 + \frac{r}{n})^{nt}$

$A(t) = 600(1 + \frac{0.02}{1})^{t}$

$A(t) = 600(1.02)^t$

Calculate the value of this investment at the end of the 5 years.

This is A(5). So

$A(5) = 600(1.02)^5 = 662.5$

The value of this investment at the end of the 5 years is of $662.5.

5 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

Perform the indicated operation

Misha Larkins [42]

Answer:

3/4

Step-by-step explanation:

1 1/4 * 3/5

Change the mixed number to an improper fraction

1 1/4 = (4*1+1)/4 = 5/4

5/4* 3/5

The 5's cancel

3/4

4 0

3 years ago

Read 2 more answers