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3 years ago

I and m are parallel to each other

Mathematics

2 answers:

Kryger [21]3 years ago

8 0

Huh? i don't understand what u mean

Misha Larkins [42]3 years ago

4 0

And what I need more information to answer your question.

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Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is g

olchik [2.2K]

If $x$ is the first number in the progression, and $r$ is the common ratio between consecutive terms, then the first four terms in the progression are

$\{x,xr,xr^2,xr^3\}$

We want to have

$\begin{cases}xr^2-x=12\\xr^3-xr=36\end{cases}$

In the second equation, we have

$xr^3-xr=xr(r^2-1)=36$

and in the first, we have

$xr^2-x=x(r^2-1)=12$

Substituting this into the second equation, we find

$xr(r^2-1)=12r=36\implies r=3$

So now we have

$\begin{cases}9x-x=12\\27x-3x=36\end{cases}\implies x=\dfrac32$

Then the four numbers are

$\left\{\dfrac32,\dfrac92,\dfrac{27}2,\dfrac{81}2\right\}$

4 0

3 years ago

olga_2 [115]

Hey! I’m not sure what you’re needing help on but if it’s an answer i’m pretty sure the answer would be - 180 (hx)

5 0

2 years ago

Solve 6 sin((π/5)x)=5 for the four smallest positive solutions

cupoosta [38]

Answer:

1.568, 3.432, 11.568, 13.432

Step-by-step explanation:

Divide equation $6\sin\left(\dfrac{\pi }{5}x\right)=5$ by 6:

$\sin\left(\dfrac{\pi }{5}x\right)=\dfrac{5}{6}.$

Then

$\dfrac{\pi }{5}x=(-1)^k\arcsin\left(\dfrac{5}{6}\right)+\pi k,\ k\in Z,$

$x=(-1)^k\dfrac{5}{\pi }\arcsin\left(\dfrac{5}{6}\right)+5k,\ k\in Z.$

Since $\arcsin\left(\dfrac{5}{6}\right)\approx 56^{\circ}\approx \dfrac{\pi }{3.2},$

four smallest positive solutions are

1. $\dfrac{5}{3.2}\approx 1.568,\ k=0;$

2. $\dfrac{-5}{3.2}+5\approx 3.432,\ k=1;$

3. $\dfrac{5}{3.2}+10\approx 11.568,\ k=2;$

4. $\dfrac{-5}{3.2}+15\approx 13.432,\ k=3.$

3 0

3 years ago

Write the next three terms in each sequence. Make sure your

vazorg [7]

Um, wouldn’t it just be 9.5, 10.6, 11.7?

8 0

3 years ago

Read 2 more answers

What is a quadratic equation? And how do you solve them?

Allisa [31]

Answer:

A quadratic equation is some thing like

(x+2)^2

(x+2)(x+5)

In order to solve them, the most basic method is the FOIL method

F-First

O-Outside

I-Inner

L-Last

So for example

(x+2)^2

Which is basically

(x+2)(x+2)

So the first is the 2 Xs

(x+2)(x+2)

x*x=x^2

The outer is the x and the 2,

(x+2)(x+2)

2*x=2x

Now the Inner which is the 2 and the x

(x+2)(x+2)

2*x=2x

And finally the last which is the 2 and the second 2

(x+2)(x+2)

2*2=4

Now add them all up

X^2+2x+2x+4

Combine like terms

x^2+4x+4

There you have it

Do the same with every other quadratic equation

3 0

2 years ago

I and m are parallel to each other​

I and m are parallel to each other