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Lynna [10]
2 years ago
12

An object is moving at a speed of 2 feet every 8.5 minutes. Express this speed in meters per hour

Mathematics
1 answer:
jekas [21]2 years ago
8 0

Answer:

4.30m/hr

Step-by-step explanation:

2 feet/8.5 mins is 0.235294118 feet per minute, which is 4.3030553875397 metres per hour, which to 2 s.f. is 4.30

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What 878.9-89.51 i need help i dont understand
Marina86 [1]

Answer: 789.39

Step-by-step explanation: Just subtract 89.51 from 878.9.

Mark Brainliest if this helped!

3 0
3 years ago
A rectangular piece of cardboard is 16y cm long and 23 cm wide. Four square pieces of cardboard whose sides are 6y cm each are c
Anna11 [10]

Answer:

Area of remaining cardboard is 224y^2 cm^2

a + b = 226

Step-by-step explanation:

The complete and correct question is;

A rectangular piece of cardboard is 16y cm long and 23y cm wide. Four square pieces of cardboard whose sides are 6y cm each are cut away from the corners. Find the area of the remaining cardboard. Express your answer in terms of y. If your answer is ay^b, then what is a+b?

Solution;

Mathematically, at any point in time

Area of the cardboard is length * width

Here, area of the total cardboard is 16y * 23y = 368y^2 cm^2

Area of the cuts;

= 4 * (6y)^2 = 4 * 36y^2 = 144y^2

The area of the remaining cardboard will be :

368y^2-144y^2

= 224y^2

Compare this with;

ay^b

a = 224, and b = 2

a + b = 224 + 2 = 226

3 0
2 years ago
HELPPPPPPPPP PLEASEEEEE<br><br> Write the perimeter of the triangle as a polynomial
GuDViN [60]
It’s either B or D. I’m sorry I’m not so sure but I think it’s either one of those.
6 0
3 years ago
A corporate team-building event costs $19plus an additional $1 per attendee. How many attendees can there be, at most, if the bu
Mars2501 [29]

Answer:

There can be at most 12 attendees in a corporate team-building event.

Step-by-step explanation:

Let x denotes number of attendees in a corporate team-building event.

Fixed cost = $19

Cost charged per attendee = $1

Budget for the corporate team-building event = $31

Therefore,

19+1(x)\leq 31\\19+x\leq 31\\x\leq 31-19\\x\leq 12

So, there can be at most 12 attendees in a corporate team-building event.

8 0
2 years ago
Can someone show how it’s done
Dmitriy789 [7]
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8 0
3 years ago
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