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nadezda [96]
3 years ago
6

The diameter of a reamed hole is 70.36 mm. What is the radius setting, in mm., used to draw the hole?

Mathematics
1 answer:
Natalija [7]3 years ago
6 0

Answer:

r = 35.18 mm

Step-by-step explanation:

We have,

The diameter of a reamed hole is 70.36 mm.

It is required to find the radius required to draw the hole.

The relation between diameter and radius is given by :

D = 2r

r is radius setting

r=\dfrac{D}{2}\\\\r=\dfrac{70.36 }{2}\\\\r=35.18\ mm

So, the radius setting used to draw the hole is 35.18 mm.

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The following set of coordinates represents which figure? (0, 4), (0, 7), (1, 5), (1, 2)
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The figure represent a parallelogram

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we have

(0, 4), (0, 7), (1, 5), (1, 2)

using a graphing tool

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see the attached figure  

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The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

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\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

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7 0
3 years ago
What is the value of a for the following circle in general form?<br> x2 + y² + ax + by+c=0
motikmotik

Answer:

Step-by-step explanation:

x2+y2+ax+by+c=0

Step 1: Add -by to both sides.

ax+by+x2+y2+c+−by=0+−by

ax+x2+y2+c=−by

Step 2: Add -x^2 to both sides.

ax+x2+y2+c+−x2=−by+−x2

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Step 4: Add -c to both sides.

ax+c+−c=−by−x2−y2+−c

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Step 5: Divide both sides by x.

ax

x

=

−by−x2−y2−c

x

a=

−by−x2−y2−c

x

y−x2

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