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2 years ago

A solid cylinder has a radius of 2 cm and a height of 7 cm. It has a density of 3.1 grams

Mathematics

1 answer:

timurjin [86]2 years ago

5 0

Answer:

Step-by-step explanation:

The first thing you need to do is find the volume.

r = 2cm

h = 7 cm

pi = 3.14

V = pi*r^2 *h

V = 3.14 * 2^2 * 7

V = 3.14 * 4 * 7

V = 87.92 cm^3

Next find the mass using the density formula

Density = mass / volume

Density = 3.1 grams / cm^3

Volume = 87.92 cm^3

3.1 = m / 87.92 Multiply both sides by 87.92

3.1 * 87.92 = m

m = 272.55

Note:

Technically There should be no decimal places. 3.1 is only 2 significant digits. The answer should be 270 grams.

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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.