Help me pls I don’t get it

Stock A changed by - $2,60 during the day. Stock B changed by 3

Slav-nsk [51]

Answer:

$7.80

Step-by-step explanation:

$2.60 * 3 = $7.80

5 0

3 years ago

X = 10 - y

victus00 [196]

I think answer should be d. Please give me brainlest let me know if it’s correct or not okay thanks bye bye

6 0

2 years ago

Kate jogged a total distance of 8 1/2 miles during the months of January and February if katie only jogged 1/4 mile everyday whi

swat32

Since in this equation we are dividing 17/2 by 1/4, we can think of this as multiplying 17/2 with 4/1 (or 4).

17 4
---- X ----- = 34
2 1

Our final answer is 34 days.

3 0

2 years ago

a football match,the ratio of children to adults is 2:7.there are 2700 people int he crowd. each adult ticket is £8 each child t

marusya05 [52]

Step-by-step explanation:

since the ratio of children to adult is 2:7,

2+7 = 9

number of children would be

(2 × 2700)/9 = 600 children in the crowd

number of adults would be

(7 × 2700)/9 = 2100 adults in the crowd

So, if a ticket costs £8 per adults, total cash made from adult

= 8 × 2100 = £16800

if a ticket costs £(8 - 3) per children, total cash made from children

= 5 × 600 = £3000

total money made = £(16800 + 3000)

= £19800

7 0

2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

2 years ago