A) Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxawould be incorrect.

ma=....

B) Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

2cos2(θ)−1 =.....

c) Enter the expression asin(x)+15, where asin(x) is the inverse sine function.

asin(x)+15 =....

D) Enter the expression (√(2gΔy)/m), where Δ is the uppercase Greek letter Delta.

(√2gΔy/m)=........

E) Enter the expression N0e−λt, where N0 is N-naught (an N with a subscript zero) and λ is the lowercase Greek letter lambda.

Respuesta :

The correct representation are as follows:

Part (a): [tex]\boxed{m\cdot a=}[/tex]

Part (b): [tex]\boxed{2*\left(\text{cos}(2\cdot \theta)\right)-1=}[/tex]

Part (c): [tex]\boxed{a*\left(\text{sin}(x)\right)+15=}[/tex]

Part (d): [tex]\boxed{\sqrt{\dfrac{2\cdot g\cdot \triangle y}{m}}=}[/tex]

Part (e): [tex]\boxed{N_{0}\cdot e^{-(\lambda\cdot t)}}[/tex]

Further explanation:

in the question it is given that while writing any mathematical expression the symbol to be used for multiplication is [tex]\cdot\text{ dot}[/tex].

For example: [tex](2\times 3)[/tex] is incorrect and [tex](2\cdot 3)[/tex] is correct.

If a mathematical expression is to be written in its explicit form then the symbol [tex]*\text{ aestrick}[/tex] is used.

Part (a):

The expression given in part (a) is as follows:

[tex]ma=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{m\cdot a=}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Part (b):

The expression given in part (b) is as follows:

[tex]2\text{cos}(2\theta)-1=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{2*\left(\text{cos}(2\cdot \theta)\right)-1=}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Part (c):

The expression given in part (c) is as follows:

[tex]a\text{sin}(x)+15=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{a*\left(\text{sin}(x)\right)+15=}[/tex]

Part (d):

The expression given in part (d) is as follows:

[tex]\sqrt{{2g\triangle y/m}=[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{\sqrt{\dfrac{2\cdot g\cdot \triangle y}{m}}=}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

Part (e):

The expression given in part (e) is as follows:

[tex]N_{0}e^{-\lambda t}[/tex]

The correct representation of the above expression is as follows:

[tex]\boxed{N_{0}\cdot e^{-(\lambda\cdot t)}}[/tex]

For multiplication [tex]\cdot\text{ dot}[/tex] symbol is used.

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Answer details:

Grade: College

Subject: Mathematics

Chapter: NA

Keywords: Expression, representation, ma, cdot, aestrick, wcos(2theta)-1, asin(x)+15, sqraureroot(2gdeltay/m), N0e-lamda t, lowercase greek, multiplication dot.

Basically, here we want to just rewrite all the multiplications using the * symbol, which represents multiplication in most common programmation languages and math calculators.

We usually do not use it when a scalar multiplies a variable, like in:

2x

but just to be complete, let's also use it as 2*x.

for example, the first expression we get is:

ma = m*a

b) Completing the others is trivial, we just need to identify where we have multiplications and add the correspondent symbol there.

[tex]2cos^2(\theta) - 1 = 2*cos^2(\theta) - 1[/tex]

c) asin(x)+15 =

Note that while "Asin(x)" is a function, I assume that here we have "a times sin(x)", so we will write:

[tex]asin(x) + 15 = a*sin(x) + 15[/tex]

D)  (√(2gΔy)/m)

Here we have the term "Δy", this is not a multiplication, this means a "difference in the value of y", so we will leave it as it is.

[tex]\frac{\sqrt{2g\Delta y} }{m} = \frac{\sqrt{2*g*\Delta y} }{m}[/tex]

E)  N0e−λt,

Here N0 is a term in itself, and actually should be written as N₀, so this is not N times zero.

So we will get:

[tex]N_0e -\lambda t = N_0*e - \lambda*t[/tex]

If instead of that, the right part was an exponent (I can't tell because of how you wrote it) we would get:

[tex]N_0e^{ -\lambda t} = N_0*e^{ - \lambda*t}[/tex]

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