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3 years ago

Which algebraic expression is equivalent to the expression below?

Mathematics

2 answers:

Paraphin [41]3 years ago

8 0

That's easy
5x+4x–x=8x

Rom4ik [11]3 years ago

5 0

Answer:
8x

Explanation:
All you have to do is simplify by combining like terms:
5x + 4x – x
9x – x
8x

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3 years ago

Explain how knowing that 5 divided by 8 is 0.625 helps you write the decimal for 4 5/8

arsen [322]

Because it become 4(0.625) then it is 2.5

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2 years ago

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

4 0

2 years ago

The distance between two villages is 0.625 km. Find the length in centimetres between the two villages on a map with a scale of

seropon [69]

Answer:

3125 cm

Step-by-step explanation:

The distance between two villages is 0.625 km. Find the length in centimetres between the two villages on a map with a scale of 1:5000.

We are given the scale of

1: 5000

This means

1 km = 5000 cm

Hence:

1 km = 5000cm

0.625 km = x cm

Cross Multiply

x cm = 0.625 × 5000 cm

x cm = 3125 cm

Therefore, the length in centimetres between the two villages is 3125cm

5 0

2 years ago

A rectangular made of square tiles measures 8 tiles wide and 10 tiles long what is the length in tiles of a similar rectangle 12

Agata [3.3K]

There’s are ratio which would be 8/10. 12/x should equally 8/10. So it would be 14 tiles long.

3 0

3 years ago