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3 years ago

A triangle height is 2/4 mm and the base is 8 mm. What is the area of the triangle?

Mathematics

2 answers:

Darina [25.2K]3 years ago

8 0

Answer:

the answer would be 2mm^2

krok68 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

Triangle Area Formula: (b*h)/2

B: 8

H: 2/4= 1/2=0.4

8*0.5/2= 2

Area ofthe triangle= 2

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Lelechka [254]

Answer:

80 lots

Step-by-step explanation:

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3 years ago

Ashley and matt each purchase one raffle ticket. If a total of 13 raffle tickets are sold and two winners will be selected,what

Naily [24]

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3 years ago

To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 55numbers (one through 55.) The ord

Romashka [77]

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3 years ago

Determine how long it will take an object to rotate through 20.0 revolutions at an angular velocity of 2.8 radians per second. U

Andreyy89

The conversion factor between radians and revolution as measures of angle is 2π rad is equal to 1 revolution. Converting the given angular velocity to rev/s,
(2.8 rad) x (1 rev / 6.28 rad) = 70/157 revolution per second
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4 years ago

LINEAR ALGEBRA

kenny6666 [7]

Answer:

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

Step-by-step explanation:

Let be $\vec u_{1} = [2,3,1]$ , $\vec u_{2} = [4,1,0]$ and $\vec u_{3} = [1, 2,k]$ , $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{3}$ if and only if:

$\alpha_{1} \cdot \vec u_{1} + \alpha_{2} \cdot \vec u_{2} +\alpha_{3}\cdot \vec u_{3} = \vec O$ (Eq. 1)

Where:

$\alpha_{1}$ , $\alpha_{2}$ , $\alpha_{3}$ - Scalar coefficients of linear combination, dimensionless.

By dividing each term by $\alpha_{3}$ :

$\lambda_{1}\cdot \vec u_{1} + \lambda_{2}\cdot \vec u_{3} = -\vec u_{3}$

$\vec u_{3}=-\lambda_{1}\cdot \vec u_{1}-\lambda_{2}\cdot \vec u_{2}$ (Eq. 2)

$\vec O$ - Zero vector, dimensionless.

And all vectors are linearly independent, meaning that at least one coefficient must be different from zero. Now we expand (Eq. 2) by direct substitution and simplify the resulting expression:

$[1,2,k] = -\lambda_{1}\cdot [2,3,1]-\lambda_{2}\cdot [4,1,0]$

$[1,2,k] = [-2\cdot\lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]$

$[0,0,0] = [-2\cdot \lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]+[-1,-2,-k]$

$[-2\cdot \lambda_{1}-4\cdot \lambda_{2}-1,-3\cdot \lambda_{1}-\lambda_{2}-2,-\lambda_{1}-k] =[0,0,0]$

The following system of linear equations is obtained:

$-2\cdot \lambda_{1}-4\cdot \lambda_{2}= 1$ (Eq. 3)

$-3\cdot \lambda_{1}-\lambda_{2}= 2$ (Eq. 4)

$-\lambda_{1}-k = 0$ (Eq. 5)

The solution of this system is:

$\lambda_{1} = -\frac{7}{10}$ , $\lambda_{2} = \frac{1}{10}$ , $k = \frac{7}{10}$

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

4 0

4 years ago