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777dan777 [17]
3 years ago
14

What is (7/12)/(5/9)? In simplest form..

Mathematics
2 answers:
kakasveta [241]3 years ago
8 0
7/12 times 5/9 equals 35/108. It is automatically in simplest form.
love history [14]3 years ago
5 0
\frac{7}{12} / \frac{5}{9} =

Change the division to multiplication through the reciprocal of the number being divided. 

\frac{7}{5} * \frac{9}{5} =  \frac{21}{20}

As mixed number:

1 \frac{1}{20}
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The area of the rectangle is 27cm2 <br> The length is 9cm<br> What is the width of the rectangle?
Luda [366]

Answer:

The width is 3cm

Step-by-step explanation:

Since the area of the rectangle is 27cm² and the length is 9cm you can just divide 27 by 9 to find out the width of the rectangle

5 0
2 years ago
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-3x - 8 + 6x = 7 wat is the answer
Aliun [14]
3x-8+6x=7
9x-8=7
9x=15
x=9/15
x=3/5

7 0
3 years ago
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Tell whether the given value is a solution of the inequality. (yes or no)<br><br> -2 ≤ k/3; k = -9
olga55 [171]

Answer:  No, it is not a solution

Work Shown:

-2 ≤ k/3

-2 ≤ -9/3

-2 ≤ -3

The last inequality is false because -3 should be smaller than -2 (not the other way around). Use a number line to help see this.

Since the last inequality is false, the original inequality must also be false for that particular k value. Therefore, k = -9 is not a solution.

4 0
2 years ago
The diameter of a semicircle is 6 miles. What is the semicircle's perimeter?
koban [17]

Answer:

15.42mi

Step-by-step explanation:

Use the formula for a semicircle's perimeter.

P=\frac{1}{2}\pi *d+d

Plug in 6 for d.

P=\frac{1}{2}\pi*6+6

Let's use 3.14 for \pi, just to make it easier, but of course, if it states to round it to something else, just plug in that many values for \pi.

P=\frac{1}{2}(3.14)*6+6 \\ \\ P=1.57*6+6 \\ \\ P=9.42+6 \\ \\ P=15.42

5 0
3 years ago
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Suppose x is a real number and epsilon &gt; 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in
kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let y\in (x-\epsilon,x+\epsilon). then we have that x-\epsilon. Hence, it makes sense to define the positive number delta as \delta=\min\{x+\epsilon-y,y-(x-\epsilon)\} (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that (y-\delta,y+\delta)\subseteq  (x-\epsilon,x+\epsilon), and if we prove this, we are done. To prove it, let z\in (y-\delta,y+\delta), then y-\delta. First, \delta \leq y-(x-\epsilon) then -\delta \geq -y+x-\epsilon hence z>y-\delta \geq x-\epsilon

On the other hand, \delta \leq x+\epsilon-y then z hence z. Combining the inequalities, we have that  x-\epsilon, therefore (y-\delta,y+\delta)\subseteq  (x-\epsilon,x+\epsilon) as required.  

3 0
3 years ago
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