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liubo4ka [24]
3 years ago
12

Can anyone help me? I don't understand this. can u also give the calculations please??

Mathematics
1 answer:
S_A_V [24]3 years ago
4 0
Is there another side to it?
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Which is equivalent to 80 1/4x
vichka [17]

Answer:

20x

Step-by-step explanation:

\frac{80}{4}=\frac{20}{1}=20

3 0
2 years ago
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The figure below shows three uniform objects: a rod with m1 = 7.00 kg, a right triangle with m2 = 3.70 kg, and a square with m3
navik [9.2K]
The answer is 

G = [m1(veca1a2) + m2 (veca3a4) + m3veca5a6)] /  (m1 + m2  + m3) 

G= m1x5 + m2 x 4sqrt (2) + m3x 3sqrt (2)  / 16  =  35 +4 x 3.7x sqrt (2) + 5.3 x3 4sqrt (2) / 16 =35 + 36<span> sqrt (2) / 16
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6 0
3 years ago
Which of the following correctly uses absolute value to show the distance between -60 and 11? (5 points)
lutik1710 [3]
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5 0
3 years ago
Question 8. PV = nRT, solve for T please?<br><br>Question 10. C= Wtc÷1000, solve for W, please?​
sp2606 [1]

Answers:

\huge \boxed{T= \frac{P V}{n R} } \\ \\ \\ \\ \huge \boxed{W= \displaystyle \frac{1000C }{tc}}

\rule[225]{225}{2}

Step-by-step explanation:

<h2>8.</h2>

P \cdot V = n \cdot R \cdot T

Dividing both sides by n · R:

\displaystyle \frac{P \cdot V}{n \cdot R} =T

<h2>9.</h2><h2 />

\displaystyle C=\frac{Wtc }{1000}

Multiplying both sides by 1000:

1000C=Wtc

Dividing both sides by tc:

\displaystyle \frac{1000C }{tc}=W

<h2 />

\rule[225]{225}{2}

4 0
3 years ago
Find the products. Show your thinking 800×500
wlad13 [49]
The product of 800x500 is 400,000
what i normally do is 8x5, which is 40.
after i get the product, i add all of the zeros.
so, the product of your question is 400,000
8 0
3 years ago
Read 2 more answers
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