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AlladinOne [14]
3 years ago
13

A little stuck on this problem help

Mathematics
1 answer:
KatRina [158]3 years ago
5 0

Answer:

∠N =  31

Step-by-step explanation:

MN = MP

∠N = ∠P = x+3

∠M + ∠N + ∠P = 180

4x+6 + x+3 + x+3 = 6x + 12 = 180

x = 28

∠N = 28 + 3 = 31

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What is the Sector area with radius 6 and an angle of 50°? Use 3.14 for pi.<br>​
nevsk [136]

Answer:

15.7 units²

Step-by-step explanation:

The sector area is 50/360 times the area of the circle (since the central area of a complete circle is 360° and that for the sector is 50°), or

Sector area = (50/360)(3.14)(6 units)² = 15.7 units²

7 0
3 years ago
Slice for 5x+c=k for x
Degger [83]

Answer:

A - x = (k-c)/5

Step-by-step explanation:

To solve this problem, we would solve for x like we would in 2x = 6 :

5x + c = k

5x = k-c (subtract c from both sides)

x = k/5 - c/5 (divide 5 on both sides)

x = (k-c)/5 (simplify)

:)

4 0
2 years ago
Read 2 more answers
What is the answer to this?
yawa3891 [41]

Answer:

5

Step-by-step explanation:

8 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
3 years ago
What is 5c-4c+c-3c and 3a+6+5a-2 and 8b+8-4b-3?
ELEN [110]

Answer: -c

8a+4

4b+5

Step-by-step explanation:

5c-4c+c-3c

c+c-3c

2c-3c

-c

3a+6+5a-2

8a+4

8b+8-4b-3

4b+5

6 0
3 years ago
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