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3 years ago

If you when your mom went to the park and there were two dogs one dog went to the owner how many dogs are at the dog park

Mathematics

2 answers:

Anna11 [10]3 years ago

4 0

Answer:

There were two dogs because the dog who went to its owner never left the park

melomori [17]3 years ago

3 0

Answer: 2 Dogs

Step-by-step explanation: The other dog never left the park

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Jordan wrote the numbers from 1 through 150. How many times did he write the digit 4 in the tens place?

olga2289 [7]

40,41,42,43,44,45,46,47,48,49,140,141,142,143,144,145,146,147,148,149

Jordan wrote the number 4 in the tens place 20 times

I hope this helps :)

6 0

3 years ago

424.25 subtract 379.4

igor_vitrenko [27]

The answer is 44.85. Hope this helped! :)

7 0

3 years ago

Read 2 more answers

Work out 12+8÷(9-5) 0.018÷0.06 Express as single fraction 5/7÷2/5

Anettt [7]

Step-by-step explanation:

I don't know if the first set of numbers is all in one set, but I'll do my best to give you an answer.

Really all you need to do is use PEMDAS for the first question.

(Parentheses, exponents, multiply, divide, add, subtract. In that order)

$1 2 + 8 \div (9 - 5) \\ 12 + 8 \div 4 \\ 12 + 2 \\ 14$

Then to simplify that fraction next to it, notice that 0.018 is 3x 0.06.

that's a 3:1 ratio, so it ends up simplifying to this:

$\frac{3}{1}$

Lastly, to solve the division of that fraction. If you divide by a fraction, you multiply whatever it's dividing by its inverse.

So...

$\frac{5}{7} \div \frac{2}{5} \\ \frac{5}{7} \times \frac{5}{2} \\ \frac{25}{14}$

7 0

3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

3 years ago

Answer A , B , C or D. Need some help!

creativ13 [48]

Answer:

Step-by-step explanation:

system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. The solution of such a system is the ordered pair that is a solution to both equations. ... The solution to the system will be in the point where the two lines intersect.

3 0

3 years ago