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3 years ago

Hey!! if you can please help me with 4 and 5 ☺️☺️

Mathematics

1 answer:

drek231 [11]3 years ago

4 0

Answer:

so sorry yarr my same question

I hope you other

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Help please asappppppppp

Aleonysh [2.5K]

Answer:

First Equation: 6a-23

Second Equation: 5a-20

Step-by-step explanation:

So, I think you want me to answer part 2...

For the first equation, we can see that the denominator stays the exact same, meaning the numerator will also not change.

For the second equation, we can see the denominator changed from a-3 to 5a-15. The denominator was simply multiplied by 5. Since the denominator was multiplied by five we must multiply the numerator by five as well.

5(a-4)

5a-20

Hope this helps :)

3 0

2 years ago

An animal shelter has 9 puppies. If the puppies are 36% of the total dog and cat population, how many dogs and cats are in the a

Luda [366]

Answer:

Let total number of dog and cat population in animal shelter be x.

As per the statement:

An animal shelter has 9 puppies.

⇒ Total number of puppies = 9

It is also given that if the puppies are 36% of the total dog and cat population.

⇒ $9 = 36\% \text{of}{\text{total number of dogs and Cat}$

$9 = \frac{36}{100} \times x$

By cross multiply we get;

$900 = 36 x$

Divide both sides by 36 we get;

$25 = x$

Therefore, total number of dogs and cat population in animal shelter are 25

5 0

3 years ago

Please help!!!! I’m about to fail and I don’t want to

adelina 88 [10]

Answer:

But, you didn't ask any question.

Step-by-step explanation:

Remember to post your question next time.

6 0

2 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers

The Barkers started their trip with a full tank of gas and a total of 39,872 miles on their car. They stopped 4 hours later and

lara [203]

Answer:

27.5 miles per gallon

Step-by-step explanation:

3 0

2 years ago