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2 years ago

7

12, 17, 22, ... Find the 36th term.

1 answer:

pychu [463]2 years ago

8 0

Answer:

187

Step-by-step explanation:

a=12,d=17-12=5

nth term =a+(n-1)d

36th term=12+(36-1)5

=12+(35×5)

12+175

36th term=187

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Can someone help me please

zvonat [6]

Step-by-step explanation:

you can follow the same procedure for others

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3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

3 years ago

Free brainless give away

Fed [463]

Answer:

Brainless bruv I gonna say that's a insult but ok

5 0

3 years ago

Read 2 more answers

A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co

butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

$\displaystyle \vec{v}=\frac{\vec{x}}{t}$

Where $\vec x$ is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

$\displaystyle \vec{x}=\vec{v}.t$

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=\ km/h$

The displacement of the plane in 2 hours is

$\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)$

$\displaystyle \vec{x_1}=km$

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=km/h$

The displacement in 1 hour is

$\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)$

$\displaystyle \vec{x_2}=km$

The total displacement is the vector sum of both

$\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+$

$\displaystyle \vec{x_t}=km$

$\displaystyle \vec{x_t}=$

The distance from the airport is the module of the displacement:

$\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}$

$\displaystyle |\vec{x_t}|=894.43\ km$

8 0

4 years ago

2) (-10) + (-7) shoe work

Sladkaya [172]

A negative plus a negative equals a negative so it's just -17

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3 years ago