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3 years ago

Please help me with this Test im really confused.

Mathematics

1 answer:

poizon [28]3 years ago

8 0

Answer:

Real Quick what is the lesson name? Dont worry tell me the lesson name and i will edit my answer to help you!

Step-by-step explanation:

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Probability and Statistics

atroni [7]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Simplify 12 to the 16th power over 12 to the 4th power

dexar [7]

$\dfrac{12^{16}}{12^4} = 12^{16 - 4} = 12^{12}$

6 0

3 years ago

Solve the equation. Then check your solution.4- 3/5(3a+4)=7A. -7.4B. -6C. 3D. -3

Evgesh-ka [11]

Answer: a = -3

Explanations:

The given equation is:

$4\text{ - }\frac{3}{5}(3a+4)\text{ = 7}$

This can be re-written as:

$4-\frac{3(3a+4)}{5}=7$

Collect like terms:

$\begin{gathered} -\frac{3(3a+4)}{5}=\text{ 7 - 4} \\ -\frac{3(3a+4)}{5}=\text{ 3} \end{gathered}$

Cross multiply:

$\begin{gathered} -3(3a+4)\text{ = 3}\times5 \\ -3(3a+4)=\text{ 15} \\ \text{Expand the bracket} \\ -9a\text{ - 12 = 15} \\ \text{Collect like terms} \\ -9a\text{ = 15 + 12} \\ -9a\text{ = 27} \\ \text{Divide both sides by -9} \\ \frac{-9a}{-9}=\frac{27}{-9} \\ a\text{ = -3} \end{gathered}$

To verify if the solution is correct, substitute a = -3 into the question given. If the Right Hand Side equals the Left Hand Side, then the solution is correct.

$\begin{gathered} 4-\frac{3(3a+4)}{5}=\text{ 7} \\ \text{Substitute a = }-3 \\ 4\text{ - }\frac{3(3(-3)+4)}{5}\text{ = 7} \\ 4\text{ - }\frac{3(-9+4)}{5}\text{ = 7} \\ 4\text{ - }\frac{3(-5)}{5}\text{ = 7} \\ 4\text{ - }\frac{-15}{5}=\text{ 7} \\ 4-(-3)=7 \\ 4+3=7 \\ 7=7 \end{gathered}$

Since the Left Hand Side = Right Hand Side = 7, the solution a = -3 is correct

5 0

1 year ago

I need help I have 20mons left

mihalych1998 [28]

Answer:

y=1/2x-7

your welcomee

8 0

3 years ago

What is the simplest form of i^16 plus i^6 minus 2i^5 plus i^13?

BARSIC [14]

$\bf \textit{recall that}\qquad i^4=1\qquad i^3=-i\qquad i^2=-1\\\\
-------------------------------\\\\
i^{16}+i^6-2i^5+i^{13}\implies i^{4\cdot 4}~+~i^{4+2}~-~2i^{3+2}~+~i^{4+4+4+1}
\\\\\\
(i^4)^4~+~i^4\cdot i^2~-~2\cdot i^3\cdot i^2~+~i^4\cdot i^4\cdot i^4\cdot i^1
\\\\\\
(1)^4+[1\cdot -1]-[2(-i)(-1)]+[(1)(1)(1)(i)]
\\\\\\
1-1-2i+i\implies -i$

6 0

3 years ago