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ZanzabumX [31]
3 years ago
10

Joeli has 10 quarters. She wants to buy postcards to mail to her friends. Each postcard costs 2 quarters. How many postcards can

she buy?
Mathematics
2 answers:
enot [183]3 years ago
6 0
5 postcards
I got the answer by dividing
hope it helped
Arada [10]3 years ago
3 0
5 post cards because its 10 divided by 2
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The value of a boat is $23,400. It loses 10% of its value every year. Find the approximate monthly percent decrease in value. Ro
Brrunno [24]

Answer:

Value of boat = $23,400

Loss of value by boat per year = 8%

To find: -  Monthly percent decrease in value of boat.  

Solution: - Decrease of value per year = 8% of $23,400 = $1,872. Monthly decrease in value = $1,872/12 = $156. Monthly percentage decrease = ($156/$23,400) * 100 = 0.6667 or 0.67 (rounded to nearest hundredth)

Step-by-step explanation:

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2 years ago
Como resuelvo esto <br>(a+2)^3​
Dmitry [639]

Answer:

a³+8

Step-by-step explanation:

4 0
3 years ago
Cosx+1/sin^3x=cscx/1-cosx
ANTONII [103]
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).

 </span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).

 </span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.

 </span>
<span>I hope this helps! </span>
8 0
3 years ago
How do u figure out my equation? I know I need to use tan, but how?
Marysya12 [62]
Basically, you need to calculate "x" correct?
The tangent of (52) = x / 20
1.2799 = x / 20
x = 1.2799 * 20
<span><span><span>x = 25.598 </span>

</span> </span>


8 0
3 years ago
Read 2 more answers
Due Soon Need Help Geometry!
AveGali [126]

 

Some basic formulas involving triangles

\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa  2 =b  2+2 + c 2

−2bc cos α

\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab   2=

 

m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m   b2 = 41(2a 2 + 2c 2-b 2)

b

Bisector formulas

\ \frac{a}{b} = \frac{m}{n}  ba =nm  

​  

 

\ l^2 = ab - mnl  2=ab-mm

A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=  

\ A = \sqrt{p(p - a)(p - b)(p - c)}A=  

p(p−a)(p−b)(p−c)

​  

 

\iits whatever  A = prA=pr with r we denote the radius of the triangle inscribed circle

\ A = \frac{abc}{4R}A=  

4R

abc

​  

 - R is the radius of the prescribed circle

\ A = \sqrt{p(p - a)(p - b)(p - c)}A=  

p(p−a)(p−b)(p−c)

​  

5 0
3 years ago
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