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3 years ago

I need help please help

Mathematics

2 answers:

IrinaK [193]3 years ago

7 0

Y=2x is the answer, its going up to and right one

Gnoma [55]3 years ago

6 0

Answer:

y=2x is the answer buddy

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<span>1/4's equivalent in lowest terms is 1/4.</span>

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Evan purchases a board game that is regularly priced at $19.95. It is on sale for 40% off. What is the sale price of the board g

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19.95 × 0.40 = $7.98.

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3 years ago

How many yellow tiles are most likely in the bag? Enter the answer in the box. A tile is randomly selected from a bag that conta

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The number of yellow tiles that are likely to be in the bag is 5.

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2 years ago

Swimming Pool On a certain hot summer's day, 236 people used the public swimming pool. The daily prices are $1.50 for children a

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Answer:

Step-by-step explanation:

$a+c=236\\ \\ a=236-c\\ \\ 1.5c+2.5a=525\\ \\ 1.5c+2.5(236-c)=525\\ \\ 1.5c-2.5c+590=525\\ \\ -c=-65\\ \\ c=65\\ \\ a=236-65\\ \\ a=171\\ \\ \text{So there were 171 adults and 65 children at the pool that day.}$

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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 $\frac{7}{10}$ .

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4 years ago