All categories

3 years ago

Help please geometry !

Mathematics

1 answer:

joja [24]3 years ago

3 0

The argument is valid by the law of syllogism

You might be interested in

If 11 runners are participating in a race, how many different ways can there be first and second place runners?

MrRa [10]

Only first and second place runners:
11*10=110 ways

Hope that helped.

6 0

3 years ago

In the table adam recorded the miles he travled each day while traveling from Florida to California. Calculate the average rate

blagie [28]

Hi!

To calculate the rate of change between days 4 and 6, we simply take those points, (4, 1503) and (6, 2196) and use the equation in the picture below:

Now, when we plug it in the equation, it looks like this:

$\frac{2196-1503}{6-4} = \frac{693}{2} = 346.5$

Therefore, your answer is 346.5 or, rounded, 347.

Meaning, your answer is C

Hope this helps and have a great day! :D

5 0

2 years ago

Which of the following statements best describes the graph of x + y = 4?

vaieri [72.5K]

The last one is the only one that will work.
For the first one the y intercept is when x = 0. The point would be 0,4

The second one gives a point of (4,0)

The third one is nonsense. (4,4) is not on this line.

4 0

3 years ago

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0

2 years ago

in the first third of the season the team played 18 games how many games will the team play during the whole season

patriot [66]

Answer:

Step-by-step explanation:

all you have to do is times 18 by 3

5 0

2 years ago

Read 2 more answers