All categories

3 years ago

If f(x) = 4x + 12 is graphed on a coordinate plane, what is the y-intercept of the graph?

Mathematics

1 answer:

Ierofanga [76]3 years ago

6 0

Answer:

it's -4 I just took the test and you will get it write

You might be interested in

Name a coordinate point located in quadrant 4

sergiy2304 [10]

A point in quadrant 4 could be (-4,-2)

3 0

3 years ago

Read 2 more answers

I need help fast I appreciate the help

Neko [114]

Answer:

3/5

Step-by-step explanation:

Sin is opp/hyp so it would be 30/50 in lowest form it would become 3/5

8 0

2 years ago

Read 2 more answers

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.