A second-degree polynomial is a quadratic polynomial. O A. True O B. False​

Answer:

the answer would be 12-19+23=18 so you need a new calculator

Step-by-step explanation:

yes

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$32000\\ r=rate\to 6.25\%\to \frac{6.25}{100}\dotfill &0.0625\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &6 \end{cases} \\\\\\ A=32000\left(1+\frac{0.0625}{1}\right)^{1\cdot 6}\implies A=32000(1.0625)^6\implies A\approx 46038.78[/tex]

Answer:

friday

Step-by-step explanation:

Answer:

11.9

Step-by-step explanation:

Making sure your calculator is in degree mode, we have:

[tex] \cos(32) = \frac{x}{14} [/tex]

[tex]x = 14 \cos(32) = 11.9[/tex]

Answer:

20%

Step-by-step explanation:

add up all the values, then do 9 (alex's score) and divide it by 45 (total that was added up)

Answer:

A. 9

Step-by-step explanation:

Add all basketball shots then divide by 5

Answer:

Formula

Volume of a rectangular prism = width × length × height

Bottom Tier

volume = 12 × 18 × 8

            = 1,728 cm³

Middle Tier

volume = 8 × 12 × 6

            = 576 cm³

Top Tier

volume = 4 × 4 × 4

            = 64 cm³

Entire Cake

Volume = bottom tier + middle tier + top tier

             = 1728 + 576 + 64

             = 2,368 cm³

Answer:

n = 20

Step-by-step explanation:

[ 180° ( n - 2 ) ] / n = 162°  

180 ( n - 2 ) = 162 n

180n - 360 = 162n

18n = 360

n = 20

Answer:

R80625

Step-by-step explanation:

Total Cost : 60000 + 4500 = 64500

To make 25% profit: 64500 X 1.25 = 80625

Answer:

Step-by-step explanation:

View the attached graph for what your image should look like.

I hope this helps or gives you a picture of what it should look like, because it's quite a long process.

Hope this helps!

Answer:

  7 m, 24 m, 25 m

Step-by-step explanation:

This problem can be solved by writing an equation expressing the given relationships and the Pythagorean theorem. Or, it can be solved by reference to common Pythagorean triples. Here, we're interested in a triple that has a difference of 1 between the hypotenuse and the longer leg. Such triples include:

  {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {9, 40, 41}, {11, 60, 61}, ...

We note that for the triple {7, 24, 25}, the longer leg is 3 more than 3 times the shorter leg: 3 +3(7) = 24.

The side lengths are:

__

In case you're unfamiliar with Pythagorean triples, or you want to write the equation, you can let s represent the length of the shorter side. Then the longer side is (3s+3) and the hypotenuse is (3s+4). The Pythagorean theorem tells you the relation is ...

  (3s +4)² = (3s +3)² +s²

  9s² +24s +16 = 9s² +18s +9 +s²

  s² -6s -7 = 0 . . . . . subtract the left side and put in standard form

  (s -7)(s +1) = 0 . . . . factor

  s = 7  or  -1 . . . . . . solutions to the equation

The side length must be positive, so the shorter leg is 7 meters long. Then the other two legs are ...

  3s +3 = 3(7) +3 = 24 . . . . meters

  3s +4 = 3(7) +4 = 25 . . . . meters

The side lengths are 7 m, 24 m, and 25 m.

Recall the geometric sum,

[tex]\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}[/tex]

It follows that

[tex]1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}[/tex]

So, we can rewrite the integral as

[tex]\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx[/tex]

Split up the integral at x = 1, and consider the latter integral,

[tex]\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx[/tex]

Substitute [tex]x\to\frac1x[/tex] to get

[tex]\displaystyle \int_0^1 \frac{\frac1{x^2} + 1}{\frac1{x^4} + \frac1{x^2} + 1} \frac{\ln\left(1 + \frac1{x^{2021}}\right) - \ln\left(1 + \frac1x\right)}{\ln\left(\frac1x\right)} \, \frac{dx}{x^2}[/tex]

Rewrite the logarithms to expand the integral as

[tex]\displaystyle - \int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2021}+1) - \ln(x^{2021}) - \ln(x+1) + \ln(x)}{\ln(x)} \, dx[/tex]

Grouping together terms in the numerator, we can write

[tex]\displaystyle -\int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2020}+1)-\ln(x+1)}{\ln(x)} \, dx + 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx[/tex]

and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to

[tex]\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx[/tex]

Substituting [tex]x\to\frac1x[/tex] again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have

[tex]\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 1010 \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 505 \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx[/tex]

We can neatly handle the remaining integral with complex residues. Consider the contour integral

[tex]\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz[/tex]

where γ is a semicircle with radius R centered at the origin, such that Im(z) ≥ 0, and the diameter corresponds to the interval [-R, R]. It's easy to show the integral over the semicircular arc vanishes as R → ∞. By the residue theorem,

[tex]\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4}\, dx = 2\pi i \sum_\zeta \mathrm{Res}\left(\frac{1+z^2}{1+z^2+z^4}, z=\zeta\right)[/tex]

where [tex]\zeta[/tex] denotes the roots of [tex]1+z^2+z^4[/tex] that lie in the interior of γ; these are [tex]\zeta=\pm\frac12+\frac{i\sqrt3}2[/tex]. Compute the residues there, and we find

[tex]\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx = \frac{2\pi}{\sqrt3}[/tex]

and so the original integral's value is

[tex]505 \times \dfrac{2\pi}{\sqrt3} = \boxed{\dfrac{1010\pi}{\sqrt3}}[/tex]

Answer:

Step-by-step explanation:

Ratio of sides of  45-45-90 triangle =  x : x : x√2

x√2 is the side opposite to angle 90

So, from the picture,

x√2 = 2√2

x = [tex]\frac{2\sqrt{2}}{\sqrt{2}}[/tex]

x = 2

Ratio of sides of 30-60-90 tirangle = a : 2a : a√3

Short side that is oppoiste to 30° = a

Side opposite to 60° = a√3

Hypotenuse (oppoiste to 90°) = 2a

a = 2

y =a√3

  y = 2√3

Answer:

D. would be the correct answer

Answer:

Step-by-step explanation:

This is a Geometric Sequence  with common ratio 15/9 = 5/3

25/15  is also =  5/3

So the 10th term = ar^(n-1)

= 9*(5/3)^9

= 893.061 to nearest thousandth.

The table represents linear because the x represent the exponential functions

Answer:

Step-by-step explanation:

Prove by multiplication

Similarly

So the expression is 5873/8

Now

We can prove the given statement to multiplication

Multiply with 8

And

Yes

We see 5873 lies between 5600 and 6400 hence the quotient is greater than 700 but less than 800

Given that the Ferris diameter is 20 meters, with a rate of rotation of 1 turn in 8 minutes, we have;

a. Amplitude: A = 10 meters

Midline: h = 11 meters

Period, P = 8 minutes

b. h(t) = 10•cos((π/4)•t + π) + 11

c. 11 meters

The amplitude is the same as the radius of the Ferris wheel,

The radius of the Ferris wheel = 20 ÷ 2 = 10

Therefore;

[tex]the \: midline \: = \frac{max \: height \: + min \: height}{2} [/tex]

Therefore;

[tex]midline \: = \frac{10 + 10 + 1 + 1}{2} = 11[/tex]

The period is the time to complete one rotation, therefore;

b. h(t) = A•cos(B•t + C) + h

Where;

B = 2•π/P

When t = 0, h(t) = 1

Which gives;

h(0) = 1 = 10 × cos(B×0 + C) + 11

-10/10 = -1 = cos(C)

C = arcos(-1) = π

Therefore;

h(0) = 1

h(0) is the minimum value of h(t)

c. After 30 minutes, we have;

h(30) = 10•cos((π/4)×30 + π) + 11 = 11

Learn more about the Ferris wheel here:

https://brainly.com/question/86214

The simulation of the medicine and the bowler hat are illustrations of probability

From the question,

From the simulation, 23 of the 30 randomly generated numbers show that the medicine is effective on at least two

So, the probability is:

p = 23/30

p = 0.767

Hence, the probability that the medicine is effective on at least two is 0.767

From the simulation, 0 of the 30 randomly generated numbers show that the medicine is effective on none

So, the probability is:

p = 0/30

p = 0

Hence, the probability that the medicine is effective on none is 0

The probability of hitting a headpin is:

p = 90%

The probability a bowler hits a headpin 4 out of 5 times is:

P(x) = nCx * p^x * (1 - p)^(n - x)

So, we have:

P(4) = 5C4 * (90%)^4 * (1 - 90%)^1

P(4) = 0.3281

Hence, the probability that the bowler hits a headpin 4 out of 5 times is 0.3281

Read more about probabilities at:

https://brainly.com/question/25870256

Answer:

the answer is 60

Step-by-step explanation:

brainliest please

Answer:

B.)   I'm not 100% sure

Step-by-step explanation:

Hope this helped

Using it's concept, it is found that the probability that the next trial will result in a 2 and 4 is given by:

A. [tex]\frac{3}{10}[/tex].

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In an experimental probability, the number of outcomes are taken from previous trials.

In this problem, the table states that of 20 trials, 6 resulted in either (2,4) or (4,2), hence the probability is given by:

p = 6/20 = 3/10.

Which means that option A is correct.

More can be learned about probabilities at https://brainly.com/question/14398287

hope it helps....!!!!!

Answer:

As Per Provided Information

An ellipse has a vertex at (0, −7), a co-vertex at (4, 0), and a center at the origin (0,0) .

We have been asked to find the equation of the ellipse in standard form .

As we know the standard equation of an ellipse with centre at the origin (0,0). Since its vertex is on y-axis

[tex] \underline\purple{\boxed{\bf \: \dfrac{ {y}^{2} }{ {a}^{2} } \: + \: \dfrac{ {x}^{2} }{ {b}^{2} } = \: 1}}[/tex]

where,

Substituting these values in the above equation and let's solve it

[tex] \qquad\sf \longrightarrow \: \dfrac{ {y}^{2} }{ {( - 7)}^{2} } \: + \dfrac{ {x}^{2} }{ {(4)}^{2} } = 1 \\ \\ \\ \qquad\sf \longrightarrow \: \dfrac{ {y}^{2} }{49} \: + \frac{ {x}^{2} }{16} = 1 \\ \\ \\ \qquad\sf \longrightarrow \: \: \dfrac{ {x}^{2} }{16} \: + \dfrac{ {y}^{2} }{49} = 1[/tex]

Therefore,

So, your answer is 2nd Picture.

Answer:

Step-by-step explanation:

We can think of this problem like a triangle that Tom, Blair, and Marcia live in.

This triangle is a right triangle.

the opposite is 3, the hypotenuse is 7, and the adjacent side is unknown.

Solving for the adjacent side will give us the distance that Tom and Blair live from each other

a^2 +b^2 = c^2

3^2 + b^2= 7^2

Now we solve for B

7^2 =49

3^3=9

9+b^2=49

Now we subtract

49-9=40

9+40=49

The square root of 40 is 6.3246

Now we check our work

3^2 +6.32^2=7^2

9+ 40 =49

The answer is, tom and Blair live approximately 6.3246 blocks from each other

Hope this helps!

Answer:

True

Determine whether x+2 is a factor of x³-3x²-7x+6:

↓

[tex]True[/tex]

I hope this helps you

:)

It looks like your question is incomplete. I believe you also have options to pick which graph is correct. However, I can still give you the information you are looking for.

The slope of the line is -1.2

The Y-intersect is (0, 4)

I have also attached an image of what the graph would look like.

Hope this helps.

Replace x with π/2 - x to get the equivalent integral

[tex]\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx[/tex]

but the integrand is even, so this is really just

[tex]\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx[/tex]

Substitute x = 1/2 arccot(u/2), which transforms the integral to

[tex]\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du[/tex]

There are lots of ways to compute this. What I did was to consider the complex contour integral

[tex]\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz[/tex]

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

[tex]\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}[/tex]

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

[tex]\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}[/tex]

and it follows that

[tex]\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}[/tex]

the ans for this q would be 5/27