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3 years ago

Suppose a student is in the last semester of college and has a 2.76 GPA after 104 credit hours. If he is

Mathematics

1 answer:

Svetach [21]3 years ago

4 0

Answer:

2.93

Step-by-step explanation:

I guessed and got it right lol

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Select all ratios equivalent to 3:5.<br> 13:30<br> 4:15<br> 6:10

uysha [10]

Answer:

6:10

Step-by-step explanation:

3:5 is equivalent to 6:10 because you multiply both 3 and 5 by 2 and you get 6:10. every thing else is wrong

6 0

2 years ago

Read 2 more answers

Solve for v.<br> -5v = -3y + 4

seropon [69]

Answer:

-5v + -3y + 4

the answer is V=3y/5 - 4/5

7 0

3 years ago

Read 2 more answers

In the star below, all of the inscribed angles are congruent. Find the measure of each inscribed angle.

ozzi

Answer:

36 degrees for each angle.

Step-by-step explanation:

If you divide 180 by 5 (for each point in the star) you get 36 degrees.

7 0

2 years ago

Which of the quadratic functions has the narrowest graph?

nadezda [96]

Hilf mir Baby, ich liebe dich

3 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