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Dmitrij [34]
2 years ago
8

In a class, 4/5 of the students have mathematical instruments. 1/4 of these students have lost their protractors. What fraction

of students in the class has protractors?​
Mathematics
2 answers:
Novay_Z [31]2 years ago
6 0

Answer:

\frac{11}{20}

Step-by-step explanation:

\frac{4}{5} - \frac{1}{4}          <em>Subtract the students who don't have protractors from the students who have mathematical instruments.</em>

\frac{11}{20}

lawyer [7]2 years ago
3 0

9514 1404 393

Answer:

  3/5

Step-by-step explanation:

We assume that every student with mathematical instruments who hasn't lost their protractor will have a protractor.

If 1/4 of the students with instruments lost their protractor, then the remaining 3/4 of students with instruments still have their protractors. The fraction of the class with protractors is ...

  (3/4)(4/5) = 3/5

3/5 of the students in the class have protractors.

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Read 2 more answers
Jenny multiplies the square root of her favorite positive integer by $\sqrt{2}$. Her product is an integer. a) Name three number
Dmitry [639]

Answer:

Part a) 2,50,18

Part b) When Jenny divides the square root of her favorite positive integer by \sqrt{2}, she gets an integer

Step-by-step explanation:

Let

x-------> the favorite positive integer

Part a)

1) For x=2

\sqrt{2}*\sqrt{2}=\sqrt{4}=2 -----> the product is an integer

so

The number x=2 could be Jenny favorite positive integer

2) For x=50

\sqrt{50}*\sqrt{2}=\sqrt{100}=10 -----> the product is an integer

so

The number x=50 could be Jenny favorite positive integer

3) For x=18

\sqrt{18}*\sqrt{2}=\sqrt{36}=6 -----> the product is an integer

so

The number x=18 could be Jenny favorite positive integer

Part B)

1) For x=2

\sqrt{2}/\sqrt{2}=\sqrt{1}=1 -----> the result is an integer

2) For x=50

\sqrt{50}/\sqrt{2}=\sqrt{25}=5 -----> the result is an integer

3) For x=18

\sqrt{18}/\sqrt{2}=\sqrt{9}=3 -----> the result is an integer

Therefore

When Jenny divides the square root of her favorite positive integer by \sqrt{2} , she gets an integer

4 0
2 years ago
In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, AE=x2−16 , and CE=6x .
MAXImum [283]

Answer:

<h3>AC=96 units.</h3>

Step-by-step explanation:

We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.

AE=x^2-16. , and CE=6x .

<em>Note: The diagonals of a parallelogram intersects at mid-point.</em>

Therefore, AE = EC.

Plugging expressions for AE and EC, we get

x^2-16=6x.

Subtracting 6x from both sides, we get

x^2-16-6x=6x-6x

x^2-6x-16=0

Factoriong quadratic by product sum rule.

We need to find the factors of -16 that add upto -6.

-16 has factors -8 and +2 that add upto -6.

Therefore, factor of x^2-6x-16=0 quadratic is (x-8)(x+2)=0

Setting each factor equal to 0 and solve for x.

x-8=0  => x=8

x+2=0  => x=-2.

We can't take x=-2 as it's a negative number.

Therefore, plugging x=8 in EC =6x, we get

EC = 6(8) = 48.

<h3>AC = AE + EC = 48+48 =96 units.</h3>
5 0
2 years ago
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