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3 years ago

Work out the following and give your answer in it's simplest form:

Mathematics

2 answers:

Pavel [41]3 years ago

8 0

Answer:

$\frac{9}{10}$ ÷ $\frac{3}{5}$

=3/5

=1.5

-BARSIC- [3]3 years ago

5 0

Here's your answer!! hopefully it's readable! have a great day :)

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Holly wants to add an odd number to 659. Which number could possibly be the sum? Choose all answers that are correct. A. 677 B.

noname [10]

It is B and C because if you add an odd number to another odd number it becomes even.

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3 years ago

CHOOSING BRAINLIEST ( SCREENSHOT PROVIDED )

Westkost [7]

Answer:

y= 25

i think

Step-by-step explanation:

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3 years ago

Find the value of 15.0 NN in pounds. Use the conversions 1slug=14.59kg1slug=14.59kg and 1ft=0.3048m1ft=0.3048m.

finlep [7]

3.37 lb

Step-by-step explanation:

The question requires you to convert weight in Newtons to weight in pounds.

Given 15.0 N to convert to pounds, remember the conversion rate where;

1 Newton = 0.224809 pound-force

1 N= 0.224809 lb

15 N= ?

Perform cross-product

=15*0.224809

=3.37 lb

Learn More

Weight conversions:brainly.com/question/762013

Keywords : value,pounds,conversions,significant figures

#LearnwithBrainly

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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

3 years ago

Does the equation 3(2x−1)+5=6(x+1) have one, none, or an infinite amount of solutions?

EleoNora [17]

Answer: No solutions

Step-by-step explanation: If you solve the problem all the way, you get 0 = 4 which is not valid so there is simply no solution

8 0

2 years ago

Read 2 more answers