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TEA [102]
3 years ago
11

What is the length of segment AE? Show your work.

Mathematics
2 answers:
MariettaO [177]3 years ago
5 0

AC+CE=AE

Pythagorean theorem is a big part of this problem. a^2+b^2=c^2

Part AC

8^2+6^2=c^2

64+36= 100

c^2 =

\sqrt{100}

or

5^2 to match c^2.

Part CE

It is similar to the other one. It's just scaled. 6 to 8, 4 to....6

4^2+6^2=c^2

16+36=52

c^2 =

\sqrt{52}

It is not a perfect square. So that will be the answer.

RSB [31]3 years ago
4 0

Answer:

Step-by-step explanation:

The short side of AB is 2 units shorter than BC.

Side DE has to be 2 units longer than side CD, which is 6.

AE will be the sum of 6 + 8 + 4 + 6, which is 24.

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What is the GCF? 2x2-4x+8
defon

Answer:

the greatest common factor is 2, so the expression could be rewritten 2(x^{2} -2x+4)

3 0
3 years ago
What is the equation of the line in slope-intercept form with slope -3/2 and y-intercept -5?
V125BC [204]

About Slope Intercept Form:

  • y = mx + b
  • m represents the slope
  • b represents the y-intercept or AKA the starting point

ABOUT PROBLEM:

  • Since -3/2 is the slope, it represents m in Slope Intercept Form
  • Since -5 is the y-intercept, it represents b in Slope Intrecept Form

y = mx + b

y = -3/2x + -5 --- IN Slope Intercept Form


Hope this helps you!!! :)


5 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
I've got a math test wish me luck​
kaheart [24]

Step-by-step explanation:

best of luck dear.........

6 0
3 years ago
2x(9-5x) - (-4x-36x)
allochka39001 [22]
I got 48x
First by distributing 2x to 9-5x=
18x-10x
Then adding a 1 in front of - making -4x-36x to 4x+36x=40x
8x+40x=48x
7 0
3 years ago
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