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3 years ago

10

What conic section is drawn by the parametric equations x=csc t and y cot?

1 answer:

Vanyuwa [196]3 years ago

3 0

If we take the square of x and square of y and then subtract them:
(csc t)²-(cot t)²=1 ( this eq. gets from basic identity
x²-y²=1......a 1+cot²x=csc²x)

equation 'a' represent the equation of hyperbola which is (x²/a²)-(y²/b²) =1 with given conditions( a=1,b=1)

So, option D is correct

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A banquet has charges $750 to feed a large party. Each person must also pay $3.50 for a tip. If divided equally, how many people

Tems11 [23]

Let the number of people be x.

Total tips = 3.5x
Total cost = 750
Each person = $15

15x = 750 + 3.5x
15x - 3.5x = 750
11.5x = 750
x = 65.22

Answer: 66 people should participate.

5 0

3 years ago

Think of 5 positive integers that have a

Rudiy27

Answer:

Mode of 3 median of 6, mean of 6, range of 8

Step-by-step explanation:

3, 3, 6, 7, 11

6 0

2 years ago

∠U=90°, TS = 73, SU = 55, and UT = 48.

ololo11 [35]

Answer:

sin S = is the ratio found.

Step-by-step explanation:

It is given that m∠U = 90°

TS is the hypotenuse = 73 units

UT is the adjacent side of the right angle = 48 units

SU is the base of the triangle = 55 units

Now we have to find the ratio as,

sin S =

sin S =

Plugin the values, we will get,

sin S =

So the ratio was found.

disclaimer this is not my work it is youaskianswer's work.

6 0

3 years ago

Prove that 1³+2³+....n³=n²(n+1)²/4. principle of mathematics induction

alex41 [277]

<h3>The Simplified Question:-</h3>

$\sf 1^3+2^3\dots n^3=\dfrac{n^2(n+1)^2}{2}$

$\\ \sf\longmapsto 1^3+2^3+\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

<h3>Solution:-</h3>

Let

$\\ \sf\longmapsto P(n)=1^3+2^3\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

For n=1

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(1+1)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(2)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{2}{2}\right)^2$

$\\ \sf\longmapsto P(1)=(1)^2$

$\\ \bf\longmapsto P(1)=1=1^3$

Let k be any positive integer.

$\\ \sf\longmapsto P(k)= 1^3+2^3\dots k^3=\left(\dfrac{k(k+1)}{2}\right)^2$

We have to prove that p(k+1) is true.

consider

$\sf 1^3+2^3\dots k^3+(k+1)^3$

$\\ \sf\longmapsto \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2}{4}+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2+4(k+1)^3}{4}$

$\\ \sf\longmapsto \dfrac{k+1)^2\left\{k^2+4k+4\right\}}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+2)^2}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+1+1)^2}{4}$

$\\ \sf\longmapsto \left(\dfrac{(k+1)(k+1+1)}{2}\right)^2$

$\\ \sf\longmapsto (1^3+2^3+3^3\dots k^3)+(k+1)^3$

Thus P(k+1) is true whenever P(k) is true.

Hence by the Principal of mathematical induction statement P(n) is true for $\bf n\epsilon N.$

Note:-

We can solve without simplifying the Question .I did it for clear steps and understanding .

<h3>Learn More:-</h3>

brainly.com/question/13253046?

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6 0

3 years ago

Read 2 more answers

23.75 into a fraction

Furkat [3]

Answer:

exact answer 91/4 the mixed fraction is 23 3/4 and then

Step-by-step explanation:

23+1/2+1/4=23 3/4

5 0

3 years ago