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3 years ago

I need some help... Algebra Hw

Mathematics

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:can't see no 1

Step-by-step explanation:

2=A

3=c

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NO LINKS PLEASE AND BE SPECIFIC i really need help!<br><br> Simplify the expression.

kondor19780726 [428]

Answer:

Step-by-step explanation:

Hope this helps!! Plz give brainliest!!!!

7 0

3 years ago

Read 2 more answers

Another story problem with fractions

ArbitrLikvidat [17]

Answer: 14 is 2 9/10 15 is 7/8

Step-by-step explanation:

Divide five and one in half, to get 2.5 and 0.5

Convert decimals to fraction form

Divide the fractions in half by finding a common multiple between the numerators

Add the numbers together and you have your answer

6 0

3 years ago

Will give brainliest to whoever answers right first

oksian1 [2.3K]

Answer:

0.5x+20=0.25x+30, m=40

Step-by-step explanation:

Blue: y=0.5x+20

Red: y=0.25+30

We’re looking for where the y (cost) is equal, so we can set the two equations to each other:

Isolate x:

0.5x+20=0.25x+30

0.5x+20-20=0.25x+30-20

0.5x=0.25x+10

0.5x-0.25x=0.25x+10-0.25x

0.25x=10

x=40

m= 40

7 0

3 years ago

How many gallons is 32 quarts

Harman [31]

4 quarts = 1 gallon
8 quarts = 2 gallons
12 quarts= 3 gallons
16 quarts = 4 gallons
20 quarts= 5 gallons
24 quarts = 6 gallons
28 quarts = 7 gallons
32 quarts = 8 gallons 8 gallons is the answer

6 0

3 years ago

Read 2 more answers

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