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3 years ago

14

En las olimpiadas matemáticas,la hora de inicio del evento se expresa en una ecuación simple: 2x+6=x+17 ¿A qué hora empezó el ev

ento?

1 answer:

Sergio [31]3 years ago

7 0

Answer:

11 minutos.

Step-by-step explanation:

En las Olimpiadas de Matemáticas, la hora de inicio del evento se expresa en una ecuación simple: 2x + 6 = x + 17 ¿A qué hora comenzó el evento?

La hora en que comienza el evento en los juegos olímpicos se representa como x

Por tanto: 2x + 6 = x + 17

Recopilar términos similares

2x - x = 17 - 6

x = 11 minutos

Por lo tanto, los eventos comienzan en 11 minutos.

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Adina has a rectangular garden that measures 9 m wide by 13 m long. She wants to

MrRissso [65]

9*13=117
12*16=192

We increased three meters to both the length and width of the garden.

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3 years ago

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Help me Solve 1/6 ÷ 2/6 please

pochemuha

$\frac{1}{6}$ ÷ $\frac{2}{6}$

First, apply this following rule: $a$ ÷ $\frac{b}{x} = a$ × $\frac{c}{b}$
$\frac{1}{6}$ × $\frac{6}{2}$
Second, apply this following rule: $\frac{a}{b}$ × $\frac{c}{d} = \frac{ac}{bd}$
$\frac{1x6}{6x2}$
Third, multiply 1 × 6 to get 6 and 6 × 2 to get 12.
$\frac{6}{12}$
Fourth, find the GCF of 6 and 12.
Factors of 6: 1, 2, 3, 6
Factors of 12: 1, 2, 3, 4, 6, 12
The GCF is 6.
Fifth, divide the numerator by the GCF.
$6$ ÷ $6 = 1$
Sixth, divide the denominator by the GCF.
$12$ ÷ $6 = 2$
Seventh, collect the new numerator and new denominator.
$\frac{1}{2}$

Answer as fraction: $\frac{1}{2}$
Answer as decimal: 0.5

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3 years ago

Subtact: (3x + 4) - (x + 2)

stiv31 [10]

Ok this is the answer to this question 2(x+1)

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3 years ago

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Sebastian was in a hotel lobby and took the elevator up 7 floors to his room. Then he took the elevator down 9 floors to the par

chubhunter [2.5K]

Answer:

7-9

Step-by-step explanation:

Sebastian goes up from 0 by 7 floors. Then, he goes down by 9.

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3 years ago

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{2})^{2}+cos^{2}(x)=1$

$cos^{2}(x)=1-\frac{1}{4}$

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

$sin(y)=\frac{\sqrt{2}}{2}$

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

3 years ago