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3 years ago

Please help please brainliest

Mathematics

2 answers:

Cerrena [4.2K]3 years ago

8 0

Answer:

i think its c sorry if am wrong

Delvig [45]3 years ago

4 0

I’m pretty sure it’s c!

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What is 12.38 in word form

cupoosta [38]

it will be twelve and thirty eight hundredths

8 0

3 years ago

Read 2 more answers

What is the nth term rule of the quadratic sequence below?

Vladimir [108]

Answer:

3n² + 5n - 2

Step-by-step explanation:

Given sequence:

6, 20, 40, 66, 98, 136, ...

Calculate the first differences between the terms:

$6 \underset{+14}{\longrightarrow} 20 \underset{+20}{\longrightarrow} 40 \underset{+26}{\longrightarrow} 66 \underset{+32}{\longrightarrow} 98 \underset{+38}{\longrightarrow} 136$

As the first differences are not the same, calculate the second differences:

$14 \underset{+6}{\longrightarrow} 20 \underset{+6}{\longrightarrow} 26 \underset{+6}{\longrightarrow} 32 \underset{+6}{\longrightarrow} 38$

As the second differences are the same, the sequence is quadratic and will contain an n² term.

The coefficient of the n² term is half of the second difference.

Therefore, the n² term is: 3n²

Compare 3n² with the given sequence:

$\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 & 3 & 12 & 27 & 48 \\\cline{1-5} \sf operation & +3&+8 & +13 & +18 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}$

The second operations are different, therefore calculate the differences between the second operations:

$3 \underset{+5}{\longrightarrow} 8 \underset{+5}{\longrightarrow} 13\underset{+5}{\longrightarrow} 18$

As the differences are the same, we need to add 5n as the second operation:

$\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 +5n & 8&22 & 42 & 68\\\cline{1-5}\sf operation & -2 &-2 &-2 & -2 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}$

Finally, we can clearly see that the operation to get from 3n² + 5n to the given sequence is to subtract 2.

Therefore, the nth term of the quadratic sequence is:

3n² + 5n - 2

6 0

2 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

3 years ago

Bianca had a weekly allowance of $8.50 two years ago. Last year, her weekly allowance was $9.75. This year, Bianca's weekly allo

Natasha_Volkova [10]

Answer:

Step-by-step explanation:

In this particular scenario, based on the numbers I would say that it does not make sense to represent this with a constant rate. That is because in a span of three years the change between each year is completely different, for example, between the first and second year there was a change of

9.75 - 8.50 = 1.25 dollar change

1.25 / 8.50 = 0.147 or a 14.7% increase

Between the second and third year, there was a change of

12 - 9.75 = 2.25 dollar change

2.25 / 9.75 = 0.23 or 23% increase

Therefore, each year the percent and dollar value increase is increasing more and more which would not be a constant rate.

3 0

3 years ago

1/9 equals what squared?

Alex

Answer:

1/3 I think??????

7 0

3 years ago