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Zina [86]
3 years ago
8

3. Approximate square root v15

Mathematics
1 answer:
baherus [9]3 years ago
4 0

Answer:

<h2>4</h2>

Step-by-step explanation:

\sqrt{15}\\=3.87298\dots \\\\= 4

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One angle of an octagon is 65º. The<br>others are equal to each other. Find them.​
r-ruslan [8.4K]

Answer:

42 1/7 degrees

Step-by-step explanation:

65+7x=360

7x=295

x=42 1/7

8 0
3 years ago
Read 2 more answers
If h and j are positive, (h+j)2 = 73 and h2 +j2 = 51, what is hj?
kkurt [141]

Answer:

hj = 11

Step-by-step explanation:

  • (h+j)^2=73,\:\: and\:\: h^2+j^2 =51 (Given)

  • (h+j)^2 = h^2+j^2 +2hj

  • \implies 73= 51 +2hj

  • \implies 73- 51 =2hj

  • \implies 22 =2hj

  • \implies hj = \frac{22}{2}

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2 years ago
A -pound bag of Feline Flavor is . An -pound bag of Kitty Kibbles is . Which statement about the unit prices is true?
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Where is the numbers dude? LOL

Step-by-step explanation:

4 0
3 years ago
Write and equation with aa variable on both sides of the equal sign that has infinitely many solutions. solve the equation and e
LekaFEV [45]
X = x

to solve, subtract both sides by x:

0 = 0

x is always equals to x so infinite solutions
8 0
4 years ago
Find a vector function, r(t), that represents the curve of intersection of the two surfaces. the paraboloid z = 9x2 y2 and the p
dolphi86 [110]

The vector function is, r(t) =  \bold{ < t,2t^2,9t^2+4t^4 > }

Given two surfaces for which the vector function corresponding to the intersection of the two need to be found.

First surface is the paraboloid, z=9x^2+y^2

Second equation is of the parabolic cylinder, y=2x^2

Now to find the intersection of these surfaces, we change these equations into its parametrical representations.

Let x = t

Then, from the equation of parabolic cylinder,  y=2t^2.

Now substituting x and y into the equation of the paraboloid, we get,

z=9t^2+(2t^2)^2 = 9t^2+4t^4

Now the vector function, r(t) = <x, y, z>

So r(t) = \bold{ < t,2t^2,9t^2+4t^4 > }

Learn more about vector functions at brainly.com/question/28479805

#SPJ4

7 0
2 years ago
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