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lakkis [162]
3 years ago
15

Matty jogs 9 km/hr. Compute Matty's speed in m/s.

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
3 0
9 km/hr=
9000m/hr=
9000m/60mins=
9000m/3600seconds=
30m/12seconds=
10m/4seconds=
5m/2seconds
2.5m/1second
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Y +2=5<br> What is the answer?
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Answer:

y = 3

Step-by-step explanation:

y + 2 = 5 \\ y = 5 - 2 \\ y = 3

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3 years ago
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A random sample of 95 vehicles is taken from a large parking lot at an office park. Below is the type of each vehicle, and wheth
kati45 [8]

Answer:

C

Step-by-step explanation:

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3 years ago
6x+6y=-6, 5x+y=-13<br>Show your process please<br>Solve the linear equation​
lisabon 2012 [21]

Answer: Rewrite equations:

5x+y=−13;6x+6y=−6

Step: Solve5x+y=−13for y:

5x+y=−13

5x+y+−5x=−13+−5x(Add -5x to both sides)

y=−5x−13

Step: Substitute−5x−13foryin6x+6y=−6:

6x+6y=−6

6x+6(−5x−13)=−6

−24x−78=−6(Simplify both sides of the equation)

−24x−78+78=−6+78(Add 78 to both sides)

−24x=72

−24x

−24

=

72

−24

(Divide both sides by -24)

x=−3

Step: Substitute−3forxiny=−5x−13:

y=−5x−13

y=(−5)(−3)−13

y=2(Simplify both sides of the equation)

Answer:

x=−3 and y=2

Step-by-step explanation:

5 0
3 years ago
Find the percent from 85 to 30​
Zanzabum

Answer:

25.5

Step-by-step explanation:

6 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
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