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2 years ago

A scuba diver descends from the surface at a rate of 38 feet per minute

Mathematics

1 answer:

Bezzdna [24]2 years ago

3 0

38+38+38=114 Hoped this helped

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The value of expression x<br> 2 + 3x<br> 2 − 5n − 2 when x = 2 is

Mumz [18]

2 + 3x

= 5

Have a Nice day

3 0

3 years ago

Which expression is equivalent to 5(x + 7)?

larisa86 [58]

You need to multiply both numbers inside the parentheses by 5:

5 times x = 5x
5 times 7 = 35

So your answer is B. 5x + 35

5 0

3 years ago

Read 2 more answers

Explain how you can use place value to describe how 0.05 and 0.005 compare

Dovator [93]

.05 is larger than .005 because larger numbers are always to the left and smaller numbers are always to the right

3 0

4 years ago

Formula for the surface area of a cylinder, SA = 2ar2 + 2arh. Solve for h.

MakcuM [25]

Answer:

$h=\frac{SA-2\pi r^2}{2\pi r}$

Step-by-step explanation:

The formula for the surface area of a cylinder is:

$SA=2\pi r^2+2\pi rh$

And we want to solve for the height h.

First, subtract $2\pi r^2$ from both sides. This yields:

$2\pi rh=SA-2\pi r^2$

Now, divide both sides by $2\pi r$ . This yields:

$h=\frac{SA-2\pi r^2}{2\pi r}$

And we're done!

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