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vitfil [10]
3 years ago
12

Explain why the exponents cannot be added in the product 12^3 and 11^3

Mathematics
1 answer:
GrogVix [38]3 years ago
4 0
Because bases are different.

a^n · a^m = a^(n+m)
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2) A cake in the shape of a circus tent is used as a centerpiece at a celebration. The cake consists of a cylinder and a cone.Th
zavuch27 [327]
<span>You have:
 
 - The diameter of the cylinder is 12 inches and its height is 14 inches.
 -The height of the cone is 6 inches.
 
 So, you must apply the formula for calculate the volume of the cylinder a the formula for calculate the volume of a cone. 
 
 V1=</span>πr²h
 <span>  
 V1 is the volume of the cylinder.
 r is the radius.
 h is the height (h=14 inches)
 
 The problem gives you the diameter, but you need the radius, so you have:
 
 r=D/2
 r=12 inches/2
 r=6 inches
 
 When you substitute the values into the formula, you obtain:
 
 V1==</span>πr²h
 V1=(3.14)(6 inches)²(14 inches)
 V1=1582.56 inches³<span>
 
 The volume of the cone is:
 
 V2=(</span>πr²h)/3
<span> 
 V2 is the volume of the cone.
 r is the radius (r=6 inches)
 h is the height of the cone (h=6 inches).
 
 Then, you have:
</span> 
 V2=(πr²h)/3
 V2=(3.14)(6 inches)²(6 inches)/3
 V2=226.08 inches³
<span> 
 Therefore, </span>the volume of the cake<span> (Vt) is:
 
 Vt=V1+V2
 Vt=</span>1582.56 inches³+226.08 inches³
<span> Vt=1808.6 inches</span>³
8 0
3 years ago
Help me please ASAP
11111nata11111 [884]

Answer:

i think it is b(-22)

Step-by-step explanation:

i hope this helps

4 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!
Softa [21]

The answer is :

( y^2 +7 )^2


8 0
3 years ago
Find the distance from P to l.
ki77a [65]
Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\

Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}
4 0
3 years ago
The value of x satisfying both the equations 4x – 5 = y and 2x – y = 3, when y = -1 is
schepotkina [342]

Answer:

x = 1

Step-by-step explanation:

4x-5=y

4x-5= -1

4x = -1+5

4x=4

x=1

2x-y=3

2x-(-1)=3

2x+1=3

2x= 3-1

2x= 2

x=1

since the two answers for x are the same therefore:

x = 1

4 0
3 years ago
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