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Slav-nsk [51]
3 years ago
6

Can someone help me answer this please

Mathematics
1 answer:
Stella [2.4K]3 years ago
7 0
It would be the third option :)
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if the radius of one of the semi-circles is 7 meters what is the circumference of one of the semi-circles​
Yanka [14]

Answer:  43.96 m

Step-by-step explanation:

the circumference of one of the semi circle radius of one of the semicircles is 7 meters is 7 meters

ur welcome

3 0
3 years ago
if two fractions have the same denominator but different numerators which fraction is greater? Give an example.
krek1111 [17]
The larger fractron will be the fraction with the greater numerator.
ex. 4/5 is greater than 2/5
7 0
3 years ago
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Keiko has a total of $ 5200 comma which she has invested in two accounts. The larger account is $ 900 greater than the smaller a
Ilia_Sergeevich [38]

Keiko has a total of $5200

x is the amount of money in larger account

y is the amount of money in smaller account

x - y = $900

and x + y = $5200

<u>This creates two simultaneous equations:</u>

x - y = $900 ... (i)

x + y = $5200 ... (ii)

Adding (i) and (ii) :

2x = $6100 , x = $3050

y = x - $900 = $3050 - $900 = $2150

The amount of money in larger account (x) = $3050

The amount of money in smaller account (y) = $2150

8 0
3 years ago
3n *power 2 +10n+7 how to factories and pls tell with solution
tankabanditka [31]

Answer:

(n + 1)(3n + 7)

Step-by-step explanation:

3n² + 10n + 7

Consider the factors of the product of the n² term and the constant term which sum to give the coefficient of the n- term.

product = 3 × 7 = 21 and sum = + 10

The factors are + 3 and + 7

Use these factors to split the n- term

3n² + 3n + 7n + 7 ( factor the first/second and third/fourth terms )

3n(n + 1) + 7(n + 1) ← factor out (n + 1) from each term

= (n + 1)(3n + 7) ← in factored form

5 0
3 years ago
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The point (-3, -5) is on the graph of a function. Which equation must be true regarding the function?
vlabodo [156]

Answer:

F(-3) = -5

Step-by-step explanation:

I got it right on the test

8 0
2 years ago
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