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3 years ago

8

Please help me!! 1st question 9−32÷4 2nd question 108÷9⋅(−12)÷6−(100÷5) 3rd question 6 + 8 ÷ 2 4th question 40÷[24−4×(2+3)] 5th

question 3u+7v+4u−2v

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

1. 1

2. -44

3. 10

4. 10

5. 7u+5v

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Which three of the following expenses can student aid cover? tuition television school supplies parties and socializing boarding

Hitman42 [59]

The tuition, school supplies, and boarding/housing for the expenses can student aid cover.

<h3>What is decision-making?</h3>

The process of making choice is by identifying the correct decision, gathering information, and assessing alternative solutions.

The following expenses can student aid cover.

A. Tuition - A student aid cover can be utilized to promote the tuition.

B. Television - There is no use of the student aid cover for selling the TV.

C. School supplies - We can use the face of the topper on the school supplies to promote schooling.

D. Parties and socializing - There is no need for student aid cover for parties and socializing.

E. Boarding or housing - We can use student aid cover to promote the hostel for the student.

More about the decision-making link is given below.

brainly.com/question/3369578

5 0

2 years ago

3x-4y=8 solve for x only<br>solve it step by step accurate answer pls <br><br>#Algebra 2

mestny [16]

Answer:

3/4x-2

Step-by-step explanation:

3−4=8

∴3=4+8

∴4=3−8

∴=14(3−8)=34−2

4 0

4 years ago

Read 2 more answers

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

What is 2 x + 2y equals negative 6 coordinate

Andrews [41]

This is solved for all values of
y = -x-3. (Assuming you meant that 2x+2y=-6).

3 0

3 years ago

PB_WITH_FSA_DIAG_6825_OL_FY21

astra-53 [7]

Answer:

uhm.....what?

Step-by-step explanation:

8 0

3 years ago