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adelina 88 [10]
3 years ago
6

Please help quick!!!!!!!!!!

Mathematics
2 answers:
larisa [96]3 years ago
8 0

Answer:

Step-by-step explanation:

Un-shaded area = 3.14 x 2 squared = 12.56

Shaded area = (3.14 x 4 squared) - Area of un-shaded area

= 50.24 - 12.56 = 37.68

Brums [2.3K]3 years ago
3 0

Answer:

37.68 sq cm

Step-by-step explanation:

Area = pi x r²

Area shaded    = area whole circle minus area white circle

pi x 4² -pi x 2²

pi x 16 - pi x4

pi(16-4)

pi(12)

3.14 x 12 = 37.68

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3 (15-6)+(18-12)^2=
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Trevor paid a total of $1.08 in fines for an overdue library book. If the penalty for overdue books is $0.09 per day, how many d
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12 because you divided 1.08 by 0.09

Step-by-step explanation:

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Unit #2 – Lesson #1 Exit Ticket: Is r=7 a solution to the equation shown below? Justify your yes no response. 6x+5=x? - 2​
Vikki [24]
(First equation:Multiply 6•7=42. 42+5=47)=(second equation:7 to the 2nd power is 49. 49-2=47 )
4 0
3 years ago
20 POINTS
RUDIKE [14]

<em>The correct expressions are as follows:</em>

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Equivalent 343

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Not Equivalent 49

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Equivalent 7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Not Equivalent 49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}

\texttt{ }

<h3>Further explanation</h3>

Let's recall following formula about Exponents and Surds:

\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }

\boxed { (a ^ b) ^ c = a ^ { b . c } }

\boxed {a ^ b \div a ^ c = a ^ { b - c } }

\boxed {\log a + \log b = \log (a \times b) }

\boxed {\log a - \log b = \log (a \div b) }

<em>Let us tackle the problem!</em>

\texttt{ }

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7^2)^{\frac{7}{5}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7)^{2\times \frac{7}{5}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5} + \frac{14}{5}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{15}{5}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{3}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{343}

\texttt{ }

<em>From the results above, it can be concluded that the correct statements are:</em>

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Equivalent 343

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Not Equivalent 49

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Equivalent 7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}

7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} Not Equivalent 49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}

\texttt{ }

<h3>Learn more</h3>
  • Coefficient of A Square Root : brainly.com/question/11337634
  • The Order of Operations : brainly.com/question/10821615
  • Write 100,000 Using Exponents : brainly.com/question/2032116

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent , Perfect , Square , Factor.

4 0
4 years ago
Read 2 more answers
HHHHHHEEEELLLLLLPPPPP
sveticcg [70]
<span>We want to check how many intersections line A and B have, that is, we want to check how many common solutions do these equations have:
</span>
i) 2x + 2y = 8

ii) x + y = 4
<span>
use equation ii) to write y in terms of x as : y=4-x, 

substitute y =4-x in equation i):

</span>2x + 2y = 8
2x + 2(4-x) = 8
<span>2x+8-2x=8
8=8

this is always true, which means the equations have infinitely many common solutions.


Answer: </span><span>There are infinitely many solutions.</span><span>

</span>
4 0
3 years ago
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