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Setler [38]
3 years ago
15

A pool has an area of 1,275 m2 and a width of 25 1/2 m. What is the length of the pool.

Mathematics
1 answer:
lisov135 [29]3 years ago
5 0

Answer:

50m is the length of the pool

Step-by-step explanation:

area= length * width

length= area/ width

1275/ 25.5 = 50

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2y + 12 = -3x<br><br> solve pls i have more I WILL GIVE EXTRA POINTS!!!!!!!!!!<br> 20+
Mrac [35]

2y + 12 = -3x

Step-by-step explanation:

Divide y y by 1 1 . Divide 12 12 by 2 2 . Use the slope-intercept form to find the slope and y-intercept. The slope-intercept form is y=mx+b y = m x + b , where m m is the slope and b b is the y-intercept.

8 0
3 years ago
Melanie swam 918 meters in 3 days if she swam the distance each day how many far did melanie swim in one day
Svetradugi [14.3K]
918/3=306
So 306 meters per day.
4 0
3 years ago
What is the height of a right circular cylinder made from a rectangular sheet of lengil
Talja [164]

Answer:

15 cm

Step-by-step explanation:

When the rectangular sheet rolled along its length, the breadth becomes the height of the cylinder, and the length becomes the circle, so the height is 15 cm

5 0
3 years ago
What expression represents the area???
galben [10]

Answer:

B.

Step-by-step explanation:

Area is width times height.

(5p-7)(3p+4)

Multiply using FOIL method

(First, outer, inner, last)

(5p*3p) + (5p*4) + (-7*3p) + (-7*4) =

15p² + 20p -21p - 28 =

15p²-p -28 =

B is the closest one. They might have made a typo.

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
3 years ago
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