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4 years ago

I need help with this , please help

Mathematics

1 answer:

PIT_PIT [208]4 years ago

3 0

Start with the 3 then move on

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express the decimal number 0.3(3 bar) in form of A/B, where A and B are intergers and B is not equal to zero

igomit [66]

x = 0.3333...

(multiply both sides by 10)

10x = 3.3333...

10x - x = 3.333... - 0.333...

9x = 3

x = 3/9 = 1/3

therefore, 0.3(3 bar) = 1/3

4 0

3 years ago

Read 2 more answers

X / 3 + 12 = 18 what is the value of x

Elanso [62]

12+3= 15 so put numbers

8 0

3 years ago

Select the system that has the same solution as -2x+2y=-5 5x- 6y =8

stiks02 [169]

Answer:

-2x+2y=-5 is equal to x=y+ 5 /2

5x- 6y =8 is equal to x= 6 /5 y+ 8 /5

Step-by-step explanation:

5 0

3 years ago

g find the 2 components of vector b = 2i + j - 3k, one parallel to a = 3i - j and another one perpendicular to a

nika2105 [10]

Answer:

The components of $\vec{b}$ parallel and perpendicular to $\vec {a}$ are $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ and $\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$ , respectively.

Step-by-step explanation:

Let be $\vec b = 2\,i+j-3\,k$ and $\vec a = 3\,i-j$ , the component of $\vec b$ parallel to $\vec a$ is calculated by the following expression:

$\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}$

Where $\hat{a}$ is the unit vector of $\vec a$ , dimensionless and $\bullet$ is the operator of scalar product.

The unit vector of $\vec a$ is:

$\hat{a} = \frac{\vec {a}}{\|\vec a\|}$

Where $\|\vec {a}\|$ is the norm of $\vec a$ , whose value is determined by Pythagorean Theorem.

The component of $\vec{b}$ parallel to $\vec {a}$ is:

$\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}$

$\|\vec {a}\| = \sqrt{10}$

$\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)$

$\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j$

$\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)$

$\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}$

$\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)$

$\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$

Now, the component of $\vec {b}$ perpendicular to $\vec{a}$ is found by vector subtraction:

$\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}$

If $\vec b = 2\,i+j-3\,k$ and $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ , then:

$\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)$

$\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$

4 0

3 years ago

Find the median of -12/4, 18/5, -25,1/5,10

sashaice [31]

Arrange from smallest to largest
-25, -12/4, 1/5, 18/5, 10
chose the middle number
1/5

4 0

3 years ago