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3 years ago

Help organize the correct order of steps THANK YOU

Mathematics

1 answer:

evablogger [386]3 years ago

6 0

E, D, B, A, C, F

I’m pretty sure that’s the order, sorry if it’s wrong

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Suppose that the local sales tax rate is 5 %

bazaltina [42]

Answer:

A: Tax paid us $1370

B: Total cost= car cost + taxes

$27,400+ $1370

=$28,770

Step-by-step explanation:

4 0

3 years ago

One of the brightest stars in the sky is Arcturus, which is part of the constellation Bootes is 32 light years from the Earth. O

larisa86 [58]

188.192 miles away.

8 0

3 years ago

Sarah sold a total of 178 t shirts and posters at a festival. She sold 46 more tshirts than poster. How many posters did she sel

ycow [4]

Sarah sold 66 posters

Solution:

Let "a" be the number of shirts sold

Let "b" be the number of posters sold

Sarah sold a total of 178 t shirts and posters at a festival

Therefore,

number of shirts sold + number of posters sold = 178

a + b = 178 ----------- eqn 1

She sold 46 more tshirts than poster

Number of shirts sold = 46 + number of posters sold

a = 46 + b --------- eqn 2

Substitute eqn 2 in eqn 1

46 + b + b = 178

2b = 178 - 46

2b = 132

b = 66

Thus she sold 66 posters

7 0

3 years ago

Thelma served five pieces of a pie. The pie was cut into eighths. What fraction of the pie did she serve? Write a multiplication

Lady bird [3.3K]

The fraction would 5/8

8 0

3 years ago

Prove that the limit x tends to 1 (2x^4-6x^3+x^2+3)÷(x-1)=-8

elixir [45]

First note that if $x\neq1$ , you have

$\dfrac{2x^4-6x^3+x^2+3}{x-1}=2x^3-4x^2-3x-3$

Now, you're looking for $\delta>0$ such that for any $\varepsilon>0$ , you have

$|x-1|$

Note that you can divide through the left side of the $\varepsilon$ inequality by $x-1$ once more:

$\dfrac{2x^3-4x^2-3x+5}{x-1}=2x^3-2x-5\implies 2x^3-4x^2-3x+5=(x-1)(2x^2-2x-5)$

So it follows that you need to find an appropriate $\delta$ that will guarantee

$|(x-1)(2x^2-2x-5)|=|x-1||2x^2-2-5|$

For the moment, let's fix $\delta=1$ . Then by this assumption, we have

$|x-1|$

From this we get

$\implies0$
$\implies0$
$\implies0$
$\implies-5$
$\implies1$

where the upper bound is what we care about. With this assumption, we then get that

$|x-1||2x^2-2-5|$

which suggests that $\delta$ can be taken to be either the smaller of 1 or $\dfrac{\varepsilon}5$ , or $\delta=\min\left\{\dfrac{\varepsilon}5,1\right\}$ , to guarantee that the function gets arbitrarily close to -8.

5 0

3 years ago