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garri49 [273]
3 years ago
5

Help ASAP pleasee!!!!

Mathematics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

243

Step-by-step explanation:

hope this helps!! god bless!! -Natalia

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I am a fraction whose Numerator and denominator are both prime numbers less than 20. If you were to increase each term by 1‚ I w
11Alexandr11 [23.1K]

Answer:

\frac{2}{5}.

Step-by-step explanation:

Let the unknown fraction be \frac{x}{y}  ,

where, x and y both are prime numbers less than 20.

Now, it is given that adding 1 to both numerator and denominator will make the fraction \frac{1}{2}.

Thus,

\frac{x+1}{y+1} =  \frac{1}{2}

2(x + 1) = y + 1

2x + 2 = y + 1

2x + 1 = y.

Clearly if x will be any odd number , two times x will be odd and adding 1 to it will result in even number and y should be even number , which is not possible as only even prime is 2.

Thus , x should be the even prime which is 2.

And y will be 5.

Thus the required fraction is \frac{2}{5}.

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3 years ago
The lengths of the sides of a triangle are in the extended ratio 7​ : 9 ​: 10. The perimeter of the triangle is 52 cm. What are
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Check the picture below

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3 years ago
Choose The Answer That Best Matches The Word In Italics.
andrey2020 [161]
<span>To be infallible means to be unable to make a mistake. In this case, selection B. Connor feels that he is able to do things without error and, in that vein, is able to beat all of his chess opponents because he knows the proper decision and moves to make at all times.</span>
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Find the series shown.
strojnjashka [21]
E^6(3n - 4)
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5 0
4 years ago
Help i’m begging plsss
olga_2 [115]

Answer:

x=8

Step-by-step explanation:

Given

AB||CD

\angle ABC = 5x + 50

\angle BCD = 7x + 34

Required

Find x

If AB||CD i.e. If AB is parallel to CD, then

\angle ABC = \angle BCD

Substitute values for ABC and BCD

5x + 50 = 7x + 34

Collect Like Terms

5x - 7x = -50 + 34

-2x = -16

Multiply through by -\frac{1}{2}

-\frac{1}{2} * -2x = -16 * -\frac{1}{2}

\frac{1}{2} * 2x = 16 * \frac{1}{2}

\frac{2x}{2} = \frac{16}{2}

x = \frac{16}{2}

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