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Inessa05 [86]
3 years ago
14

How can I make 8 and 62/40 a proper mixed number

Mathematics
2 answers:
Lina20 [59]3 years ago
6 0
You can't make it proper.You can only simplify.
Step 1: Factor both the numerator and denominator down to prime factors:<span>62 = 2 * 31 </span>
40 = 23<span> * 5</span>Step 2: Calculate the greatest common factor, GCF (also called greatest common divisor, GCD), by taking all the common prime factors of the numerator and denominator, by the lowest powers:gcd(2 * 31; 23<span> * 5) = </span><span>2 </span>
Step 3: Divide both the numerator and denominator by their greatest common factor, GCF (also called the greatest common divisor, GCD):62/40 =
(2 * 31)/<span>(23 * 5)</span> = 
<span>((2 * 31) : 2)</span> / <span>((23 * 5) : 2)</span> =
31/<span>(22 * 5)</span> =
31/20Step 4: Improper fraction, rewrite:<span>31 : 20 = 1 and remainder = 11 => </span>
31/20<span> = </span>(1 * 20 + 11)<span> / </span>20<span> = 1 + </span>11/20<span> = 1 </span>11/20<span> as a </span>mixed number (also called mixed fraction).<span>1 + 11 : 20 = 1.55 as a </span>decimal number<span>.</span>
slavikrds [6]3 years ago
3 0
First you change 62/40 to a proper mixed number: 1 11/20

then you add 8 to 1 11/20 :

your final answer is 9 11/20


hope this helps :)


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please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me
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Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are  equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH=\sqrt{(5-0)^{2}+(1-1)^{2}}=5 and

GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5

Now, mid point of JH=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

=(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)

Mid point of GI=(\frac{5}{2},1)

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

(GI)^{2}=(IJ)^{2}+(JG)^{2}                             (1)

Now, JI=\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20} and GJ=\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}

Substituting these values in (1), we get

5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }

25=20+5

25=25

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5  and HJ=\sqrt{(0-5)^2+(1-1)^2}=5 are equal, thus, GHIJ is a rectangle.

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