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Reil [10]
2 years ago
12

!!!!! determine the function being differentiated, and the number at which its derivative is being evaluated. Where possible, ev

aluate the limits using differentiation.

Mathematics
1 answer:
Genrish500 [490]2 years ago
8 0

Recall that the derivative of a function f(x) at a point x = c is given by

\displaystyle f'(c) = \lim_{x\to c} \frac{f(x) - f(c)}{x - c}

By substituting h = x - c, we have the equivalent expression

\displaystyle f'(c) = \lim_{h\to0} \frac{f(c+h) - f(c)}h

since if x approaches c, then h = x - c approaches c - c = 0.

The two given limits strongly resemble what we have here, so it's just a matter of identifying the f(x) and c.

For the first limit,

\displaystyle \lim_{h\to0} \frac{\sin\left(\frac\pi3 + h\right) - \frac{\sqrt3}2}h

recall that sin(π/3) = √3/2. Then c = π/3 and f(x) = sin(x), and the limit is equal to the derivative of sin(x) at x = π/3. We have

(\sin(x))' = \cos(x)

and cos(π/3) = 1/2.

For the second limit,

\displaystyle \lim_{a\to0} \frac{e^{2a} - 1}a

we observe that e²ˣ = 1 if x = 0. So this limit is the derivative of e²ˣ at x = 0. We have

\left(e^{2x}\right)' = e^{2x} (2x)' = 2e^{2x}

and 2e⁰ = 2.

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Simplify<br> X2-49<br> Over <br> X2+9x+14
Stels [109]

Answer:

(x-7)/(x+2)

Step-by-step explanation:

(x-7)(x+7)/(x+7)(x+2)

(x-7)/(x+2)

7 0
3 years ago
A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns​ shown, making the​
Ilya [14]

Answer:

The original selling price would be $ 515.87 ( approx )

Step-by-step explanation:

Let x be the original selling price ( in dollars ),

After marking down 10%,

New selling price = x - 10% of x = x - 0.1x = 0.9x

Again after marking down 30%,

Final selling price = 0.9x - 30% of 0.9x

= 0.9x - 0.3 × 0.9x

= 0.9x - 0.27x

= 0.63x

According to the question,

0.63x = 325

\implies x =\frac{325}{0.63}\approx 515.87

Hence, the original selling price would be $ 515.87.

7 0
2 years ago
A is an m×n matrix.Check the true statements below:A. The kernel of a linear transformation is a vector space.B. If the equation
Bess [88]

Answer:

Results are (1) True. (2) False. (3) False. (4) True. (5) True. (6) True.

Step-by-step explanation:

Given A is an m\times n matrix.  Let T :U\to V  be the corresponding linear transformationover the field F and \theta be identity vector in V. Now if x\in Ker( T)\implies T(x)=\theta.

(1) The kernel of a linear transformation is a vector space : True.

Let x,y\in Ker( T), then,

T(x+y)=T(x)+T(y)=\theta+\theta=\theta\impies x+y\in Ker( T)

hence the kernel is closed under addition.

Let \lambda\in F, x\in Ker( T), then

T(\lambda x)=\lambda T(x)=\lambda\times \theta=\theta

\lambda x\in Ker(T) and thus Ket(T) is closed under multiplication

Finally, fore all vectors u\in U,

T(\theta)=T(\theta+(-\theta))=T(\theta)+T(-\theta)=T(\theta)-T(\theta)=\theta

\implies \theta\in Ker(T)

Thus Ker(T) is a subspace.

(2) If the equation Ax=b is consistent, then Col(A) is \mathbb R^m : False

if the equation Ax=b is consistent, then Col(A) must be consistent for all b.

(3) The null space of an mxn matrix is in \mathbb R^m

: False

The null space that is dimension of solution space of an m x n matrix is always in \mathbb R^n.

(4) The column space of A is the range of the mapping x\to Ax

: True.

(5) Col(A) is the set of all vectors that can be written as Ax for some x. : True.

Here Ax will give a linear combination of column of A as a weights of x.

(6) The null space of A is the solution set of the equation Ax=0.

: True

5 0
3 years ago
Leah's parents are planning to give her a gift and want to wrap it first. The gift is in a rectangular box with a height of 3 in
jekas [21]
128 square inches of wrapping paper, since the width would be the Z Axis, That would make a Square 3X4= 12, 12X2= 24. 3X7= 21X4= 104, 104+24=128? That's what I figured. Can you send me a photo of the problem?
8 0
3 years ago
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
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