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Ilia_Sergeevich [38]
3 years ago
8

begin by dividing 2/3 by 4 then add 1/4 to result multiply that result by 72 divide that result by 1/6

Mathematics
1 answer:
Aleksandr [31]3 years ago
5 0
179.9712 is your answer i believe although that is an obscure equation
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A line passes through (-9,-5) and (2,-4)
zlopas [31]

The point-slope form of a line:

y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-9, -5) and (2, -4). Substitute:

m=\dfrac{-4-(-5)}{2-(-9)}=\dfrac{1}{11}\\\\y-(-5)=\dfrac{1}{11}(x-(-9))\\\\\boxed{y+5=\dfrac{1}{11}(x+9)}\leftarrow\boxed{\text{point-slope form}}

The standard form of a line:

Ax+By=C

y+5=\dfrac{1}{11}(x+9)                <em>multiply both sides by 11</em>

11y+55=1(x+9)

11y+55=x+9     <em>subtract 9 from both sides</em>

11y+46=x      <em>subtract 11y from both sides</em>

46=x-11y

\boxed{x-11y=46}\leftarrow\boxed{\text{standard form}}

5 0
3 years ago
Need help with my math models homework
valentinak56 [21]

Answer:

you haven't put down a photo

5 0
4 years ago
How many faces,edges,and vertices does this have
alukav5142 [94]
Faces- 5
Edges- 9
Verticies- 6
7 0
3 years ago
Read 2 more answers
I need help it's geometry
nekit [7.7K]

Step-by-step explanation:

\sin(45)  =  \frac{x}{5}  \\   \frac{1}{ \sqrt{2} } =  \frac{x}{5}    \\ x =  \frac{5}{ \sqrt{2} }  = 3.535

\sin(45)  =  \frac{y}{5}  \\ y =  \frac{5}{ \sqrt{2} }  = 3.535

3 0
3 years ago
Read 2 more answers
Each CD is approximately 4.72 inches in circumference and 0.05 inches in depth. There are 15 CDs in the stack. Use the Cavalieri
Tju [1.3M]

Answer:

SA = 7.08 in²

Step-by-step explanation:

We can use the following method to solve the given problem

SA = 2B + LA

Where LA

LA = 4.72* 0.05* 15= 3.54

And

C = 2πr

Where

C = 4.72in

π = 3.14

Substituting, we have

4.72 = 2*3.14*r

Making r the subject of formula

r = 0.75in

Solve for the area of CD

A = πr²

Substituting the values we have

A = 3.14 *(0.75)²

A= 1.77 in²

Solving for surface area of the stack

SA = 2(1.77) + 3.54

SA = 7.08 in²

8 0
4 years ago
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