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Ira Lisetskai [31]
3 years ago
10

In a circle, the circumference and diameter vary directly. Which of the following equations would allow you to find the diameter

of a circle with a circumference of 154 if you know that in a second circle the diameter is 14 when the circumference is 44? A.154d = (14)(44) B.154/d = 44/14 C.14d = 154/44
Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
8 0
The answer is <span>B.154/d = 44/14 
</span>
<span>In a circle, the circumference (C) and diameter (D) vary directly:
C = kD
k = C/D

Circle 1: k = C1/D1
Circle 2: k = C2/D2</span>

C1/D1 = C2/D2
C1 = 154
D1 = ?
C2 = 44
D2 = 14

154/D1 = 44/14
So, the choice B. is correct.
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A group of friends wants to go to the amusement park. They have no more than $150
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Step-by-step explanation:

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2 years ago
I need help 6x-2y=10 x-2y=-5 solve by elimination
dem82 [27]
<h3><u>Explanation</u></h3>
  • Given the system of equations.

\begin{cases} 6x - 2y = 10 \\ x - 2y =  - 5 \end{cases}

  • Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

\begin{cases} 6x - 2y = 10 \\  - x  +  2y =  5 \end{cases}

There as we can get rid of the y-term by adding both equations.

(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x =  \frac{15}{5}  \longrightarrow  \frac{ \cancel{15}}{ \cancel{5}}  =  \frac{3}{1}  \\ x = 3

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

x - 2y =  - 5 \\ 3 - 2y =  - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\  \frac{8}{2}  = y \\ y =  \frac{8}{2} \longrightarrow  \frac{ \cancel{8}}{ \cancel{2}}  =  \frac{4}{1}  \\ y = 4

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

  • Answer Check by substituting both x and y values in both equations.

<u>First</u><u> </u><u>Equation</u>

6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10  = 10 \longrightarrow \sf{true} \:  \green{ \checkmark}

<u>Second</u><u> </u><u>Equation</u>

x - 2y =  - 5 \\ 3 - 2(4) =  - 5 \\ 3 - 8 =  - 5 \\  - 5 =  - 5 \longrightarrow  \sf{true} \:  \green{ \checkmark}

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3><u>Answer</u></h3>

\begin{cases} x = 3 \\ y = 4 \end{cases} \\  \sf \underline{Coordinate \:  \: Form} \\ (3,4)

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i saed to answer it and it didnt show me the image so im conf

Step-by-step explanation:

so probably they are, unlees one is acxute and the other is obtuse then.... ur on ur own buddy,

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