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-Dominant- [34]
4 years ago
7

A large grape fruit has a diameter of 12cm. What is the radius?

Mathematics
2 answers:
lyudmila [28]4 years ago
6 0
6 cm 
the diameter is two times the radius so the radius would be 12/2 which is 
6 cm 
trapecia [35]4 years ago
3 0
I believe its 1.25 but do some research 

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In 1910, the population of Seattle, Washington, was 237,194. In 2010, the population had grown to 608,660. By about how much did
Wewaii [24]
You subtrscy the bigger number (2010's population) by the smaller one (1910's population). 237194-608660= 371,466.

When you round to the 100's you check the 10's place. If 5 or higher add a 1 to the hundreds place, and make the 1's and 10's place 0. If lower don't add anything and make the 1's and 10's place into 0s.

When you round it, you will get 371,400
7 0
3 years ago
I need help please and explain me thanks
erastova [34]
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5 0
3 years ago
Read 2 more answers
What does the product of any whole-number factor multiplied by 100 always have ? Explain
Arisa [49]
<span>The product of any whole number factor multiplied by 100 will always have two zeros on the end, or a zero as the digit in both the tens and units column. For example, 1 x 100 is 100, 46 x 100 is 4600 and 7258 x 100 is 725,800.</span>
6 0
3 years ago
Given: ∆ABC, m∠C = 90° CB = 8, m∠B = 38º Find the area of a circumscribed circle. Find the area of the inscribed circle.
vitfil [10]

Answer:

Circumscribed circle: Around 80.95

Inscribed circle: Around 3.298

Step-by-step explanation:

Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:

\cos 38= \dfrac{8}{AB} \\\\\\AB=\dfrac{8}{\cos 38}\approx 10.152

To find the area of the circumscribed circle:

r=\dfrac{AB}{2}\approx 5.076 \\\\\\A=\pi r^2\approx 80.95

To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:

AC=\sqrt{10.152^2-8^2}\approx 6.25

The area of the triangle is:

A=\dfrac{bh}{2}=\dfrac{8\cdot 6.25}{2}=25

The semiperimeter of the triangle is:

\dfrac{10.152+6.25+8}{2}\approx 24.4

The radius of the circle is therefore \dfrac{25}{24.4}\approx 1.025

The area of the inscribed circle then is \pi\cdot (1.025)^2\approx 3.298.

Hope this helps!

6 0
3 years ago
A rectangle with length n and height m. A right triangle with hypotenuse n + 1, length n, and height m.
Oksana_A [137]

9514 1404 393

Answer:

  C. 12cm

Step-by-step explanation:

The equation for the perimeter of the rectangle is ...

  P = 2(L+W)

  34 = 2(n +m)

Solving for m, we get

  m = 17 -n . . . . . . . divide by 2, subtract n

__

The Pythagorean theorem gives the relationship between the sides and the hypotenuse

  m^2 +n^2 = (n+1)^2

  (17 -n)^2 +n^2 = (n +1)^2 . . . . . . substitute for m

  289 -34n +n^2 +n^2 = n^2 +2n +1 . . . . eliminate parentheses

  n^2 -36n +288 = 0 . . . . . . . put in standard form

  (n -12)(n -24) = 0 . . . . . . . . . factor

  n = 12 . . . . . . . . . . n=24 is an extraneous solution here

The value of n is 12 cm.

8 0
3 years ago
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