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bija089 [108]
3 years ago
6

Jayson used 0.5 cup of kernels to make 16 cups of popcorn. How many cups of popcorn would he make if he used 0.75 cup of kernels

?
Mathematics
2 answers:
Tresset [83]3 years ago
7 0

Answer:28

Step-by-step explanation:

Hope I helped

Brainliest plz?

Ber [7]3 years ago
4 0

Answer:

24 cups

Step-by-step explanation:

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H = -16t2 + 64t + 80 Determine the maximum height of the rocket. Show your work.
Lesechka [4]
H = -16t² + 64t + 80 (:16)
H = -t² + 4t + 5
/\ = b²-4ac
/\ = 4²-4.(-1).5
/\ = 16+20
/\ = 36
Yv = -/\ / 4a
Yv = -36/4.(-1)
Yv = -36/-4
Yv = 9
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I need help on number 2 quickkkkk!!!
Svetlanka [38]

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80

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4 years ago
Whats the answer to this equation 3g+5=17
Papessa [141]
3g+5=17 \ \ |-5\\ \\3g+5-5=17-5\\ \\3g=12 \ \ /:3 \\ \\g=\frac{12}{3}\\ \\ g=4


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3 years ago
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Radio stations use electromagnetic waves for broadcasting. The chart shows different frequencies of waves used by radio stations
Neko [114]

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Step-by-step explanation:

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3 years ago
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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
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