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klemol [59]
3 years ago
9

PLEASE HELP!! GRADE 7 MATHEMATICS!!!

Mathematics
1 answer:
Rzqust [24]3 years ago
5 0
The area of 3x5 face is used 4 times rather then twice
S=94in ^2
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central angle of a sector is 72° and the sector has an area of 16.2(pie) ft^2. what is the radius of the circle?
Cloud [144]
Sector area = (central angle / 360) * PI * radius^2
sector area = (72 / 360) * PI * radius^2
radius^2 = sector area / [(72 / 360) * PI]
radius^2 = 16.2 * PI / [(1 / 5) * PI]
radius^2 = 16.2 / .2
radius^2 = 81
radius = 9

Source:
http://www.1728.org/radians.htm



6 0
3 years ago
PLEASE HELP I CANT FIGURE THEM OUT
statuscvo [17]

Answer:they all equal 69

Step-by-step explanation:

6 0
3 years ago
9. The expression -8x? – 56x written in fully factored form is
Scorpion4ik [409]

Answer:

the answer is C

Step-by-step explanation:

hope this helps!

4 0
2 years ago
This is a question from my sibling's math class.
Solnce55 [7]

Answer:

14

Step-by-step explanation:

You can solve using PEMDAS:

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

3(2+5)-5(3)+8=?

3(7)-5(3)+8=?

No exponents

3(7)-5(3)+8=?

21-15+8=?

No division

21-15+8=?

21-7=?

21-7=14

3(2+5)-5(3)+8=14

5 0
2 years ago
Read 2 more answers
The figure here shows triangle AOC inscribed in the region cut from the parabola y=x^2 by the line y=a^2. Find the limit of the
aleksandrvk [35]
Area of the parabolic region = Integral of [a^2 - x^2 ]dx | from - a to a =

(a^2)x - (x^3)/3 | from - a to a = (a^2)(a) - (a^3)/3 - (a^2)(-a) + (-a^3)/3 =

= 2a^3 - 2(a^3)/3 = [4/3](a^3)

Area of the triangle = [1/2]base*height = [1/2](2a)(a)^2 = <span>a^3

ratio area of the triangle / area of the parabolic region = a^3 / {[4/3](a^3)} =

Limit of </span><span><span>a^3 / {[4/3](a^3)} </span>as a -> 0 = 1 /(4/3) = 4/3
</span>
 



3 0
3 years ago
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