Help me please I need help.

gayaneshka [121]

Answer:

8977

Step-by-step explanation:

Hello,

$49+50+51+52+...+141+142\\\\=(48+1)+(48+2)+(48+3)+(48+4)+...+(48+93)+(48+94)\\\\=48\times 94+(1+2+3+4+...+93+94)\\\\=48\times 94 + \dfrac{94\times 95}{2}\\\\=47(48\times 2+95)=47\times 191=8977$

thanks

7 0

2 years ago

Find the circumference of the circle. Show your work , round to the nearest 10th .

tester [92]

Answer:

C = 144.4 yds

Step-by-step explanation:

C = 2πr Use the equation of circumference and substitute the given radius

C = 2(3.14)(23) Multiply

C = 144.4 yds

4 0

2 years ago

HELP ME AGAIN I MIGHT PASS

trapecia [35]

Answer:

x=70

Step-by-step explanation:

it's 70 because 50 and 20 is proportional to x

4 0

2 years ago

Read 2 more answers

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

2 years ago

8th term of 1/4,1/8,1/16,1/32

sukhopar [10]

$a_1=\dfrac{1}{4};\ a_2=\dfrac{1}{8};\ a_3=\dfrac{1}{16};\ a_4=\dfrac{1}{32};...\\\\it's\ a\ geometric\ sequence\ where\ a_1=\dfrac{1}{4}\ and\ r=\dfrac{1}{8}:\dfrac{1}{4}=\dfrac{1}{2}\\\\a_n=a_1r^{n-1}\\\\n=8\\\\a_8=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)^{7-1}=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{4}\cdot\dfrac{1}{64}=\dfrac{1}{256}$

4 0

2 years ago