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3 years ago

Which net represents the figure? AKA the rectangle

Mathematics

1 answer:

alexira [117]3 years ago

4 0

Https://us-static.z-dn.net/files/d97/2509b700ec60779b64191f8a268eeaba.png

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A conical perfume bottle has a radius of 3.3 centimeters and a height of 3.3 centimeters.

expeople1 [14]

Answer:

4.5

Step-by-step explanation:

7 0

3 years ago

Graph 7x+6y<42 7x+6y>-42

suter [353]

7x + 6y < 42
6y < -7x + 42
y < -7/6x + 7

slope = -7/6....y int = (0,7)...x int = (6,0)...it will be a dashed line...shading will be below the line....so start at (0,7)...and since the slope is -7/6...go down 7 spaces and to the right 6 spaces, and keep doing this and u will cross the x axis at (6,0)

7x + 6y > - 42
6y > -7x - 42
y > -7/6x - 7

slope = -7/6...y int = (0,-7)....x int = (-6,0)....line will be dashed....shading will be above the line...so start at (0,-7) and since the slope is -7/6...go down 7 spaces and to the right 6 spaces...keep doing this and u will cross the x axis at (-6,0)

5 0

3 years ago

The mean, the median, and the mode of a set of numbers can sometimes be equal to each other.

yaroslaw [1]

Answer:

True.

Step-by-step explanation:

A data set can have the same mean, median, and mode.

3 0

4 years ago

Read 2 more answers

The enrollment at a local college increased by 9% over last year's enrollment of 3900. Find the current enrollment.

lubasha [3.4K]

Increased enrollment =9%
Last enrollment=3900
Current enrollment=9%of3900+3900
=9×3900/100+3900
=351+3900
=4451

6 0

3 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