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OleMash [197]
3 years ago
15

46 degrees and x X =

Mathematics
2 answers:
Alika [10]3 years ago
4 0

Answer:

x = 136°

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

x is an exterior angle of the triangle , then

x = 90° + 46° = 136°

vesna_86 [32]3 years ago
3 0

Answer:

136° that's 90 plus 46.........

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True or false this number is written in scientific notation? <br> .2486 x 10^12
Scilla [17]

Answer:

<h2>It Is True </h2>

It is true because scientific notation is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 ✕ 10^8.

8 0
3 years ago
Simplify the product using FOIL. (3x + 4)(2x – 6)
Arturiano [62]

Answer:

6x^2 - 10x - 24

Step-by-step explanation:

  1. Do 3x times 2x which is 6x^2
  2. Do 3x times - 6 which is -18x
  3. Do 4 times 2x which is 8x
  4. Do 4 times - 6 which is - 24
  5. You get 6x^2 - 18x + 8x - 24
  6. Simplyify to get the answer
5 0
4 years ago
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What is the surface area of the rectangular<br> prism shown by the net?
brilliants [131]

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80

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There are four 4 by 3 rectangles, so I did 4x3 and got 12, then multiplied by 4 to get 48.  There are also two 4 by 4 squares, so 4x4=16 and 16x2=32

32+48=80

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3 years ago
Historically, the average time to service a customer complaint has been 3 days and the standard deviation has been 0.50 day. Man
kicyunya [14]
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6 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
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