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timofeeve [1]
3 years ago
10

Un número consta de dos cifras cuya suma es 9.Si se invierten el orden de las cifras el resultado es igual al número dado más 9

unidades. Encuentra dicho número.
Mathematics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

45

Step-by-step explanation:

A number consists of 2 digits whose sum is 9 => x + y = 9

If the order of the numbers is inverted, the result is equal to the given number plus 9 units => (10*y + x) + 9 = (10*x + y)

Thus:

x + y = 9  => multiply by -9 => 9*y + 9*x = 81

10*y + x + 9 = 10*x + y =>  9*y - 9*x = -9

We are left then that:

9*y + 9*x = 81  (1)

9*y - 9*x = -9  (2)

we add both equations

18y = 72

y = 72/18 = 4

replacing

x + y = 9

x + 4 = 9

x = 5

Therefore the number is (10y + x) => (10 * 4 + 5) => 45

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Solve for f (-3) for f (x) = 1 -x<br> f (-3) = [ ? ]
Lostsunrise [7]
F(x) = 1 - x
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What type of triangle is formed by joining the points D(7, 3), E(8, 1), and
KiRa [710]

D(7, 3), E(8, 1), and  F(4, -1)

We calculate the squared distances between the points:

DE^2 = 1^2 + 2^2 = 5

EF^2=4^2+2^2=20

DF^2=3^2+4^2=25

Since DF^2=DE^2+EF^2 we have a right triangle with hypotenuse DF.

7 0
4 years ago
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
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dem82 [27]

Answer:

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Step-by-step explanation:

All you have to do is multiply the numerator and denominator with the same number. For example, I did 2. 4x2=8, and 2x13=26.

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