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makkiz [27]
2 years ago
5

Keith is riding an ferris wheel.His car turned 120 degrees and stopped .Then his car turn 100 degrees and stop again how many de

grees did the car turned in all
Mathematics
1 answer:
vesna_86 [32]2 years ago
5 0

Answer:

220 degrees

Step-by-step explanation:

120 degrees + 100 degrees= 220 degrees

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The numbers -9 and -4 add up to -13 and multiply to get 36

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Solve the system of linear equations by substitution 2x-y=6 x=y-1
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substitute x for y-1
2(y-1)-y=6 distribute
2y-2-y=6 subtract y in 2y
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A drug prescription of 45 pills costs $427. 95. If each pill contains 3 mg of medication, what is the cost per milligrams?
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2 years ago
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Eric noticed that his basil plant grows 5 cm every day. He starts recording the height when the plant is 22 cm tall. He wants to
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The answer is B because you multiply 14 days (2 weeks) to 5 which is 70 then you add the other 22cm and that's how you would end up with 92cm
5 0
3 years ago
hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and
Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
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