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Sladkaya [172]
3 years ago
11

Find the perimeter of this shape​

Mathematics
1 answer:
Margaret [11]3 years ago
4 0
The answer is 257.08m. Explanation: you know that one side of the square in the middle is 50m. This is the same as the diameter of one of the half circles on each side. The perimeter of a circle is the formula C=2(pi)r, where r is the radius. Dividing 50 by two will get you the size of the radius, then just plug it into the equation to find the perimeter of both rounded sides (a full circle). The two circular sides equal 157.08m, so then you just need to add the flat sides, which are both 50m since they’re congruent. 157.08m + 50m + 50m = 257.08m, so the perimeter of the shape is 257.08m.
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Can someone please explain how to figure this out?!!​
Keith_Richards [23]
It’s basically the opposite just switch it and put the 5/t now
4 0
3 years ago
What is the product in lowest terms?
Alisiya [41]
-5/12 × 8/13

First, we need to start out by applying the multiplication rule towards fractions. If you haven't learned or don't remember the rules, you can always look up fraction rules. The rule is: a/b × c/d = ac/bd. Basically, we are combining the numerators and both of the denominators.
- \frac{5 \times 8}{12 \times 13}

Second, let's now multiply what we have in the fraction. (5 × 8 = 40) and (12 × 13 = 156). Doing so will create a new fraction for us to use.
-\frac{40}{156}

Third, now obviously, we can simplify the fraction we just got into lower terms. To do that, we have to collect the greatest common factor (GCF) of both the numerator and denominator and list their factors to find the common factor that is the greatest.

Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40
Factors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156

Out of the listed factors, which of those are the common ones? 1, 2, and 4 are the common factors. Since we are looking for the greatest common factor, it would be 4 since that is the higher number out of the commons. The GCF is 4.

Fourth, we can now divide our numerator (40) and our denominator (156) by the GCF we just found out, which was 4. 
40 \div 4 = 10  \\ 156 \div 4 = 39

Fifth, our last step is to create our new simplified fraction. All we have to do is take our new numerator and denominator to see the fraction. 

Answer in fraction form: \fbox {-10/39}
Answer in decimal form: \fbox {-0.2564}
4 0
3 years ago
Prove: The square of a number that is
Nikolay [14]

Let 3<em>n</em> + 1 denote the "number" in question. The claim is that

(3<em>n</em> + 1)² = 3<em>m</em> + 1

for some integer <em>m</em>.

Now,

(3<em>n</em> + 1)² = (3<em>n</em>)² + 2 (3<em>n</em>) + 1²

… = 9<em>n</em>² + 6<em>n</em> + 1

… = 3<em>n</em> (3<em>n</em> + 2) + 1

… = 3<em>m</em> + 1

where we take <em>m</em> = <em>n</em> (3<em>n</em> + 2).

3 0
2 years ago
Please explain how to solve this (<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7Ba%5E5%20b%5E3%7D%20" id="TexFormula1" ti
olchik [2.2K]
2 is the base and (\sqrt[4][a^5 b^3]) is the exponent.

the 4 is called an index and it means to fi d the fourth root of the expression under the (\sqrt[][text]) symbol.\\a fourth root is a factor that was multiplied four times. ex: 2*2*2*2=16. the fourth root of 16 is two. \\when you do square roots, you take the number representing the factor out of the radical. anything else that is not a square root stays under. ex: \\([tex]\sqrt[9][text])\\([tex]\sqrt[3*3*3][text]) there is one set of two 3's so\\([tex]3\sqrt[3][text])\\with a fourth root, you look for groups of four. the symbol of that group is placed outside the radical. anything else stays in.\\ ([tex]\sqrt[4][a^5 b^3])

(\sqrt[4][a*a*a*a*a*b*b*b]) there is one group of 4 a's so

(a\sqrt[4][a* b^3])

there really isn't anything else to do to simplify the expression
5 0
3 years ago
Janet and Jack are walking in a two-day event to raise money for charity. Janet asks a friend for a $7.50 donation, plus $0.10 f
il63 [147K]

Answer:

Step-by-step explanation:

A. 40

B. 28

C. 44

D. 25

3 0
3 years ago
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