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hram777 [196]
3 years ago
15

Write an equality with the solution x <12

Mathematics
1 answer:
Tom [10]3 years ago
4 0
Lol i learned this last chapter. ok here is one 2x + 3 < 5
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Explain how you would write 423,090,709,000 in word form
vfiekz [6]
Four hundred twenty-three billion, ninety million, seven hundred nine thousand


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8 0
3 years ago
the cost of a phone call lasting 3 minutes 30 seconds was 52.5 cents . at this rate what was the cost of a call lasting 5 minute
Effectus [21]

Answer:

80 cents

Step-by-step explanation:

The easiest place to start for this is to calculate how much it costs per minute of call time. To do this, if we know that it costs 52.5 cents to call for 3.5 minutes, we can divide those two numbers to get how much it costs per minute.

52.5/3.5 = 15

If it costs 15 cents per minute, and we want to know how much it would cost to call for 5.33 (5 and 1/3 of a minute), then we multiply our 15 cents a minute by the number of minutes to get the final cost.

15 x 5.33 = 79.99

Because we can't have 99/100 cents, rounding up to 80 is important to get a proper answer.

6 0
3 years ago
I need help ASAP!!!!!!!!!!
Lorico [155]

Answer:

00

Step-by-step explanation:

7 0
3 years ago
How do you solve inequalites
zhenek [66]
<span>Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.But these things will change direction of the inequality: ...<span>Don't multiply or divide by a variable (unless you know it is always positive or always negative)</span></span>
7 0
3 years ago
Let f (x) = 3x − 1 and ε &gt; 0. Find a δ &gt; 0 such that 0 &lt; ∣x − 5∣ &lt; δ implies ∣f (x) − 14∣ &lt; 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
2 years ago
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