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Orlov [11]
2 years ago
7

Nate solves the equation 3*5=15 by writing and solving 15:5=3. Explain why Nate's method works

Mathematics
2 answers:
nirvana33 [79]2 years ago
8 0
Because the ratio 15 to 5 is 3 .  5 goes into 15 3 times. 
Talja [164]2 years ago
4 0
Nate's method works because division is the opposite of multiplication. Nate just checked to see if 15/5 was three, which is the same as 3x5 being 15
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2 years ago
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Medians $\overline{dp}$ and $\overline{eq}$ of $\triangle def$ are perpendicular. if $dp= 18$ and $eq = 24$, then what is ${de}$
monitta
Denote by M the point of  intersection of the medians. 
Denote also the distance DM by x and the distance QM by y. 
From the median properties of triangles we know that
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Also, 
x=\frac{2}{3}\times DP=\frac{2}{3}\times 18=12
Since the medians are perpendicular, we deduce that:
x^2+y^2=DQ^2\iff DQ=\sqrt{8^2+12^2}=14.4
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2 years ago
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The following system of equations is _____.
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First equation: y = -1/3x + 2/3
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In short, The following system of equation is "Equivalent"

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2 years ago
The price of a box of 15 cloud markers is $12.70. The price of a box of 42 cloud markers is $31.60. All prices are without tax,
stich3 [128]

Answer:  

Here, The price of one box having N markers = Price of N markers + packaging charge

Let M be the price of one Marker and P' be the price of packaging of one box which is same for any size.

Since, The price of a box of 15 cloud markers is $12.70.

That is , 15 M + P' = 12.70 ---------(1)

Also,  The price of a box of 42 cloud markers is $31.60,

That is, 42 M + P' = 31.60 ---------(2),

Equation (2) - Equation (1),

27 M = 18.90,

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By putting the value of P in equation (1),

We get,

10.5 + P' = 12.70 ⇒ P' = 2.2,

Thus, the price of one marker, M = 0.7

And, the price of packaging one box, P' = 2.2

Thus, the price of a box of 50 marker = 50 × M + P' = 50×0.7 + 2.2 = 35 + 2.2 = $ 37.2

And, the equation that shows the price(P) of a box of N markers,

P = 0.7 N + 2.2

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2 years ago
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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating
MakcuM [25]

Answer:

A=8.4063u^{2}

Step-by-step explanation:

Be the functions:

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according the graph:

\int\limits^1_7 {\frac{3}{x} } \, dx -\int\limits^1_7 {\frac{3}{x^{2} } } \, dx =3\int\limits^1_7 {\frac{1}{x} } \, dx -3\int\limits^1_7 {\frac{1}{x^{2} } } \, dx=3(\int\limits^1_7 {\frac{1}{x} } \, dx -\int\limits^1_7 {\frac{1}{x^{2} } } \, dx)=3[lnx-\frac{1}{x}](1-7)=3[(ln7-ln1)-(\frac{1}{7}-1)]=3[(1.945-0)-(0.1428-1)]=3*(1.945+0.8571)=3*2.8021=8.4063u^{2}

6 0
2 years ago
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