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Anni [7]
4 years ago
10

Find the value of x in the equation x⁄ 4 + x⁄14 + x⁄17 = 71

Mathematics
1 answer:
Morgarella [4.7K]4 years ago
4 0

Hello from MrBillDoesMath!

Answer:

33769/181

Discussion:

Each term contains "x" so factoring it out gives

x( 1/4 + 1/14 + 1/17)  = 71       (*)

Use common factor (17*14*4  = 952) as the denominator to combine terms:

1/4 =  (17*14)/ 952  = 238/952

1/14 = (17*4)/952 = 68/952

1/17 = (14*4)/952  = 56/952

so 1/4 + 1/14 + 1/17 =    (238 + 68 + 56)/ 952 =  362/952 = 181/476

Substituting in (*) gives

x ( 181/476)  = 71                       => multiply both sides by 476/181

x = (71 * 476)/181                       => 71* 476 =33769

x = 33769/181

Thank you,

MrB

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Answer:

83% would be 0.83 as a decimal

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3 years ago
A car drives with an average velocity of 20.1 m/s . After some time, it has traveled 34.2 km. How long has the car been travelin
exis [7]

Answer:

time = distance / velocity

time = 34,200 meters / 20.1 meters per second

time = 1,701.4925373134 seconds

time = 28.36 minutes

 

Step-by-step explanation:

3 0
3 years ago
Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.
alina1380 [7]

Answer:

v_1=(\frac{1}{10},-\frac{3}{10})

v_2=(-\frac{1}{10},\frac{3}{10})

Step-by-step explanation:

First we define two generic vectors in our \mathbb{R}^2 space:

  1. v_1 = (x_1,y_1)
  2. v_2 = (x_2,y_2)

By definition we know that Euclidean norm on an 2-dimensional Euclidean space \mathbb{R}^2 is:

\left \| v \right \|= \sqrt{x^2+y^2}

Also we know that the inner product in \mathbb{R}^2 space is defined as:

v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

\left \| v_1 \right \|= \sqrt{x^2+y^2}=1

and

\left \| v_2 \right \|= \sqrt{x^2+y^2}=1

As second condition we have that:

v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0

v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0

Which is the same:

y_1=-3x_1\\y_2=-3x_2

Replacing the second condition on the first condition we have:

\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}

Since x_1^2= \frac{1}{10} we have two posible solutions, x_1=\frac{1}{10} or x_1=-\frac{1}{10}. If we choose x_1=\frac{1}{10}, we can choose next the other solution for x_2.

Remembering,

y_1=-3x_1\\y_2=-3x_2

The two vectors we are looking for are:

v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})

5 0
3 years ago
Find the equation of (2,-1)
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You seem to have forgotten to add in the y-intercept or slope. For the information you have given, it could be any equation, so long as it passes through the point (2, -1), such as
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Is it A, B, or C? PLEASE HELP
Zanzabum

Step-by-step explanation:

y = kx, where k is the constant of proportionality. (1)

Therefore y = x and line B depicts that.

6 0
3 years ago
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