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Stella [2.4K]
3 years ago
8

Oh god please help I have no idea what this is​

Mathematics
2 answers:
ycow [4]3 years ago
6 0

Answer:

1. 10/7, 2/1, 1,5

2. 5/8 * 3/2

Step-by-step explanation:

(:

ehidna [41]3 years ago
5 0

Answer:

Which part exactly.. Or which number exactly

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Composite shapes find the total volume of the 3D shapes round to the nearest hundredth
Nat2105 [25]

Answer:

904.32inc^3

Step-by-step explanation:

<em>S</em><em>E</em><em>E</em><em> </em><em>I</em><em>M</em><em>A</em><em>G</em><em>E</em><em> </em><em>F</em><em>O</em><em>R</em><em> </em><em>A</em><em>N</em><em>S</em><em>W</em><em>E</em><em>R</em><em>.</em><em>.</em><em>.</em>

8 0
3 years ago
- can somebody please help me .
Dominik [7]
21. 0.25x - 1.25 - 3x + 2 = -2.75x + 0.75
22. 15/100= 0.15. 0.15•64,000= 9,600. The answer is D.
23. 1,200-300= 9,000. 1,200+300= 15,000. 9,000/15,000= 3/5. 3/5= 0.6. 60% of the building is aboveground.
24. The patterson seems to be the y=18x and when you plug 7 in to the equation, you get 126, not 128, so D is the answer.
3 0
3 years ago
I don't get this I really need help to understand this "/
Alisiya [41]
I don't either it is quite hard to be fair
4 0
4 years ago
Can someone please help me answer my question
algol [13]
The surface area would be 150cm and the volume would be 125cm. To get volume you do 5 cubed to get the answer 125. To get surface area you would do 5 multiplied by 5 multiplied by the cubes 6 sides to get 150.
4 0
3 years ago
Read 2 more answers
A rectangle has a perimeter of 30 feet and an area of 50 square feet. What are the dimensions of the rectangle?
REY [17]
Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):
a = l \times w \\ l =  a \div w \\ p = 2l + 2w = 2(l + w)
since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:
p = 2(l + w) = 2((a \div w) + w) \\  = 2((a \div w) + ( {w}^{2} \div w)) \\  p= 2((a +  {w}^{2}) \div w
Now plug in p and a values:
p= 2((a +  {w}^{2}) \div w) \\ 30 = 2((50 +  {w}^{2}) \div w)  \\ 30 \div 2 = (50 +  {w}^{2}) \div w \\ 15w = 50 +  {w}^{2}
15w = 50 +  {w}^{2}  \\  {w}^{2}  - 15w + 50 = 0 \\ (w - 5)(w - 10) = 0
therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10

D) 5 feet, and 10 feet
3 0
3 years ago
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