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AlladinOne [14]
3 years ago
6

Can someone explain how to write a function with this table, and assist me in writing one?

Mathematics
1 answer:
schepotkina [342]3 years ago
8 0
Wow.
Usually this type of tables are using for lines. But your case is more specific. I suppose that you just mixed up two last coordinates. If so, here is an answer:

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A cone with a diameter of 12 inches has a volume of 640.88 cubic inches. find the height of the cone
BlackZzzverrR [31]

Answer:

The height of the cone is 17\ in

Step-by-step explanation:

we know that

The volume of a cone is equal to

V=\frac{1}{3}\pi r^{2}h

In this problem we have

r=12/2=6\ in ----> the radius is half the diameter

V=640.88\ in^{3}

assume \pi=3.14

substitute the values and solve for h

640.88=\frac{1}{3}(3.14)(6)^{2}h

h=640.88*3/((3.14)(36))=17\ in

8 0
3 years ago
5 + 3(2+6) ÷ 4 <br> can someone explain how to solve this
konstantin123 [22]

Answer: 4.5

Step-by-step explanation:

do the numbers that are in parentheses first which is 2+6=8 next is 5+3=8 then add both ur answers together then divide that by 4 which should equal 4.5

7 0
3 years ago
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i have to write equations in standard form using integer coefficients for A,B, and, C Example: y= -8/15x + 1/20
uysha [10]

Answer:

c

Step-by-step explanation:

5 0
3 years ago
find the midpoint of the line segment whose endpoints are (-2, 5) and (4, -9). (2, -7) (1, -4) (1, -2)
Alex_Xolod [135]
M(x, y) = ((x1 + x2)/2, (y1 + y2)/2) = ((-2 + 4)/2, (5 - 9)/2) = (2/2, -4/2) = (1, -2)
7 0
3 years ago
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PLEASE help me with this question and I WILL love you forever But yeah please help me with the question.And EXPLAIN in words how
Rufina [12.5K]
Well on this problem recall area of a rectangle is defined by:

A = lw    our w = 1/6 and our l = (x + \frac{2}{3})   we have let 'x' represent our original length.
(x +  \frac{2}{3} )( \frac{1}{6} )= \frac{11}{72}
\frac{1}{6} x+ \frac{2}{18} = \frac{11}{72}
12x + 8 = 11
12x = 11 - 8
12x = 3
x = \frac{3}{12} = \frac{1}{4}  So our original length is \frac{1}{4}
5 0
3 years ago
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