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Gnom [1K]
2 years ago
14

Can you help me out here im in the test rn

Mathematics
2 answers:
iris [78.8K]2 years ago
4 0
I pretty sure it’s D
Fofino [41]2 years ago
3 0

Answer:

Type of exercise

Step-by-step explanation:

You might be interested in
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
Write equation for the ellipse with foci (9,0) and (-9,0) and endpoints on the verticle axis of (0,2) and (0,-2).
Inga [223]
See the answer on the attached 

6 0
3 years ago
3(-4/7) I don’t know how to get the answer
WARRIOR [948]

Answer:

Step-by-step explanation:

if its 3 - 4/7 then it = 2 3/7

if its 3*-4/7 then it = -12/7 = - 1 5/7

7 0
3 years ago
Ok ok this i need help
lakkis [162]

Answer:

2

Step-by-step explanation:

18-4^2=2

7 0
2 years ago
Read 2 more answers
Verify seca(1-sina)(seca+tana)=1
beks73 [17]
Seca(1-sina)(seca+tana)=1
Left hand side
=(1-sina) (seca +tana) / cosa
=seca +tana - sinaseca - sinatan / cosa
=seca + tana -tana -(sin^2a/cosa) / cosa
=(1/cosa - sin^2a/cosa) / cosa
= (1-sin^2a / cosa) / cosa
= (1-sin^2a)/ (cos^2a)
=1 (verified)
5 0
3 years ago
Read 2 more answers
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