You can make an equation: 6.75 + x = 10
6.75 is the amount you spend on the book, x is the amount you can spend on the drink, and 10 is the total cost.

Solve for x to find how much you can spend on the drink.
6.75 + x = 10
Subtract 6.75 from both sides to isolate x.

x = 3.25

You can spend $3.25 on the drink.

6 0

2 years ago

Find the inequality represented by the graph pleasseee help me

gayaneshka [121]

Answer:

y≥1/2x is the answer for this problem

8 0

2 years ago

Why do I think I'm ugly

Tanya [424]

Answer:

Ok listen here

Step-by-step explanation:

You are like the earth.

On the inside is a bright center and lava. When lava comes to the surface as a volcano, many people think it's ugly and dangerous, but other people think that it's beautiful.

We all know that a lot of people think the surface of the earth is ugly, and they have given up hope that people can change it, but the truth is that, the world is beautiful and we can change it.

You are beautiful and the people on your earth are represented as your brain. Your brain needs to tell you to look at yourself and to say that you are beautiful. You can change your mindset to tell you that you don't care what other people say, all that matters is what you say.

Always remember that :)

7 0

2 years ago

Read 2 more answers

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

2 years ago