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2 years ago

9

I really need sum help plz

1 answer:

timama [110]2 years ago

3 0

Answer:

32 Strawberries

Step-by-step explanation:

The amount of strawberries you need for 48 cups of juice is 32

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Show the working out too please in numbers, thank you.

vazorg [7]

Hii :))

$\tt \: - 5 = \frac{x}{6} \\ \tt \: - 5 \times 6 = x \\ \boxed{ \tt \: - 30 = x}$

__________________

The correct value of x is <u>-</u><u> </u><u>3</u><u>0</u><u>.</u>

$\overbrace{ \underbrace{ \sf \: \infty \: Carry \: on \: Learning \: \infty }}$

ᴛʜᴇᴇxᴛʀᴀᴛᴇʀᴇꜱᴛʀɪᴀʟ

3 0

2 years ago

Read 2 more answers

A class of 40 students has 11 students who ride the bus and 10 students who are in the band. Three of the students who ride the

Alexxandr [17]

The denominator is 40
10-3=7 (number of students in the band who do not ride the bus)
7/40=0.175
0.174*100=17.5% chance
Hope this helps!

8 0

3 years ago

PLEASE HELP!! ILL MARK YOU BRAINLIEST IF YOU ANSWER THE FASTEST AND CORRECTLY :) 10 POINTS

Ostrovityanka [42]

Answer:

Here is yours.

But next time, pls give it a try,bro!

Good luck!

3 0

3 years ago

What is the height of a cylinder with a volume of 384 pi cubic inches and a radius of 8 inches? Round to the nearest tenth of an

Nataliya [291]

Answer: 1.9 inches

Step-by-step explanation:

- the equation to find the volume of a cylinder is V = π r^2 h

- so we can just change this equation around to H = V / π r^2

- so H = 384/ π 8^2 = 1.91 or 1.9

hope this helps :)

8 0

1 year ago

A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

mestny [16]

Answer:

The value is $W= 2640 \ ft \cdot lb$

Step-by-step explanation:

From the question we are told that

The weight of the bucket is $F = 5 lb$

The depth of the well is $x_1 = 60 \ ft$

The weight of the water is $W_w = 42 lb$

The rate at which the bucket with water is pulled is $v = 1.5 \ ft/s$

The rate of the leak is $r = 0.15 lb/s$

Generally the workdone is mathematically represented as

$W = \int\limits^{x_1}_{x_o} {G(x)} \, dx$ ]

Here G(x) is a function defining the weight of the system (water and bucket ) and it is mathematically represented as

$G(x) = F + (W_w- Ix)$

Here I is the rate of water loss in lb/ft mathematically represented as

$I = \frac{r}{v}$

=> $I = \frac{0.15 }{1.5 }$

=> $I = 0.1$

So

$G(x) = 5 + (42- 0.1x)$

=> $G(x) = 47- 0.1x)$

So

$W = \int\limits^{60}_{0} {47- 0.1x} \, dx$ ]

=> $W = [47x - \frac{0.1x^2}{2} ]|\left 60} \atop {0}} \right.$

=> $W= [47(60) - 0.05(60)^2]$

=> $W= 2640 \ ft \cdot lb$

7 0

3 years ago