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Stor (34 20 +27 +21 +29+19)Step 2: 34 - 25 =9 20-25=-5
27 - 25= 2 21 - 25 = -4
29 - 25 = 4 19 - 25 = -6
Step 3: 9+ (-5) + 2 + (-4) + 4+ (-6)=0
Step 4:
O = 0

Answers

Answer 1
Answer:

Answer:

what is the question???

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14V + 5 − 5V = 4(V + 15)

Which statement describes this system of equations?
9x – 6y = 15,
3x – 2y = 5

Answers

This system of equation has an infinite number of solutions.

You can tell this because multiplying one equation by a constant gives you exactly the other equation. You can multiply the second equation by 3.

9x - 6y = 15

5(3x - 2y = 5) = 9x - 6y = 15

When this happens, you can assume that there are an infinite number of solutions.

Answer:

1) The equations in the system are equivalent equations.

2) There is no solution to the system of equations.

3) The system of equations has one solution at (3, 2).

4) The system of equations has one solution at (5, 5).

Answer = 1

Step-by-step explanation:

I just did the assignment

Please Explain/Show your work!

Answers

Answer:

  24 crates

Step-by-step explanation:

The usable load capacity for 40-kg crates is ...

  1050 kg - 82 kg = 968 kg

The number of 40-kg crates that can be loaded in a trailer with this capacity is ...

  floor(968/40) = floor(24.2) = 24 . . . . crates

_____

The function floor(n) gives the greatest integer less than or equal to n. For positive numbers, it drops the fraction.

A new DVD is available for sale in a store one week after its release. The cumulative revenue, $R, from sales of the DVD in this store in week t after its release is R=f(t)=350 ln tR=f(t)=350lnt with t>1. Find f(5), f'(5), and the relative rate of change f'/f at t=5. Interpret your answers in terms of revenue.

Answers

Solution :

It is given that :

$f'(t) = (350 \ln t)'$

       $=350(\ln t)'$

        $=(350)/(t)$

So, f(5)=350 \ln (5) \approx 563

     $f'(5) = (350)/(5)$

              =70

The relative change is then,

$(f'(5))/(f(5))=(70)/(350\ \ln(5))$

         $=(1)/(5\ \ln(5))$

         $\approx 0.12$

          =12\%

This means that after 5 weeks, the revenue from the DVD sales in $563 with a rate of change of $70 per week and the increasing at a continuous rate of 12% per week.

Consider the optimization problem where A m × n , m ≥ n , and b m . a. Show that the objective function for this problem is a quadratic function, and write down the gradient and Hessian of this quadratic.

b. Write down the fixed-step-size gradient algorithm for solving this optimization problem.

c. Suppose that Find the largest range of values for α such that the algorithm in part b converges to the solution of the problem.

Answers

Answer:

Answer for the question :

Consider the optimization problem where A m × n , m ≥ n , and b m .

a. Show that the objective function for this problem is a quadratic function, and write down the gradient and Hessian of this quadratic.

b. Write down the fixed-step-size gradient algorithm for solving this optimization problem.

c. Suppose that Find the largest range of values for α such that the algorithm in part b converges to the solution of the problem.

is explained din the attachment.

Step-by-step explanation:

H(x) = −(x − 5)2 + 3g(x) = 4 cos(2x − π) − 2
f(x)

x y
0 −5
1 0
2 3
3 4
4 3
5 0
6 −5
Which function has the largest maximum?

Answers:
 f(x)
 g(x)
 h(x) 
All three functions have the same maximum value.

Answers

Answer:

The correct option is 1. The function f(x) has the largest maximum.

Step-by-step explanation:

The vertex form of a parabola is

y=a(x-h)^2+k

Where, (h,k) is vertex.

The given functions is

h(x)=-(x-5)^2+3

Here, a=-1, h=-5 and k=3. Since the value of a is negative, therefore it is an downward parabola and vertex is the point of maxima.

Thus the maximum value of the function h(x) is 3.

g(x)=4\cos (2x-\pi)-2

The value of cosine function lies between -1 to 1.

-1\leq \cos (2x-\pi)\leq 1

Multiply 4 on each side.

-4\leq 4\cos (2x-\pi)\leq 4

Subtract 2 from each side.

-4-2\leq 4\cos (2x-\pi)-2\leq 4-2

-6\leq 4\cos (2x-\pi)-2\leq 2

Therefore the maximum value of the function g(x) is 2.

From the given table it is clear that the maximum value of the function f(x) is 4 at x=3.

Since the function f(x) has the largest maximum, therefore the correct option is 1.
Max(f(x)) = f(3) = 4

Max(g(x)) = g(π/2) = 2

Max(h(x)) = h(5) = 3

Of these values, f(3) = 4 is the largest.


The function with the largest maximum is f(x).

Prove the trigonometric identity
(tan x + cot x)/(csc x * cos x) = sec^2 x​

Answers

Answer:

(\tan x + \cot x)/(\csc x \cos x)=\sec^2 x

\boxed{((\sin x)/(\cos x) + (\cos x)/(\sin x))/((1)/(\sin x) \cdot \cos x)}=\sec^2 x

\boxed{((\sin^2 x)/(\sin x\cos x) + (\cos^2 x)/(\sin x \cos x))/((\cos x)/(\sin x))}=\sec^2 x

\boxed{((\sin^2 x+\cos^2 x)/(\sin x\cos x))/((\cos x)/(\sin x))}=\sec^2 x

\boxed{((1)/(\sin x\cos x))/((\cos x)/(\sin x))}=\sec^2 x

\boxed{(1)/(\sin x\cos x) \cdot (\sin x)/(\cos x)}=\sec^2 x

(1)/(\cos^2x)=\sec^2x

\sec^2x=\sec^2x

Step-by-step explanation:

Given trigonometric identity:

(\tan x + \cot x)/(\csc x \cos x)=\sec^2 x

\textsf{Use the identities\;\;$\tan x = (\sin x)/(\cos x)$\;,\;$\cot x=(\cos x)/(\sin x)$\;\;and\;\;$\csc x=(1)/(\sin x)$}:

\boxed{((\sin x)/(\cos x) + (\cos x)/(\sin x))/((1)/(\sin x) \cdot \cos x)}=\sec^2 x

Simplify the denominator and make the fractions in the numerator like fractions:

\boxed{((\sin^2 x)/(\sin x\cos x) + (\cos^2 x)/(\sin x \cos x))/((\cos x)/(\sin x))}=\sec^2 x

\textsf{Apply\;the\;fraction\;rule\;\;$(a)/(b)+(c)/(b)=(a+c)/(b)$\;to\;the\;numerator}:

\boxed{((\sin^2 x+\cos^2 x)/(\sin x\cos x))/((\cos x)/(\sin x))}=\sec^2 x

\textsf{Use\;the\;identity\;\;$\sin^2x+\cos^2x=1$}:

\boxed{((1)/(\sin x\cos x))/((\cos x)/(\sin x))}=\sec^2 x

\textsf{Apply\;the\;fraction\;rule\;\;$(a)/((b)/(c))=a \cdot (c)/(b)$}:

\boxed{(1)/(\sin x\cos x) \cdot (\sin x)/(\cos x)}=\sec^2 x

Cancel the common factor sin x, and apply the exponent rule aa = a² to the denominator:

(1)/(\cos^2x)=\sec^2x

\textsf{Use the identity\;\;$(1)/(\cos x)=\sec x$}:

\sec^2x=\sec^2x

Answer:

The proof of the trigonometric identity:

We can start by expanding the numerator and denominator. In the numerator, we can use the trigonometric identities tan x = sin x / cos x and cot x = cos x / sin x.

In the denominator, we can use the trigonometric identity csc x = 1 / sin x. This gives us:

((tan x + cot x))/((csc x * cos x) ) = (((sin x )/( cos x)) + ((cos x )/(sin x)))/(((1)/( sin x)) * cos x)

`We can then cancel the sin x terms in the numerator and denominator. This gives us:

((tan x + cot x))/((csc x * cos x) ) = (1 + 1)/(((1 )/(sin x)) * cos x)

We can then multiply the numerator and denominator by sin x. This gives us:

((tan x + cot x))/((csc x * cos x) ) = (sin x + sin x)/((1 )/(cos x))

We can then simplify the expression. This gives us:

((tan x + cot x))/((csc x * cos x) ) = (2sin x)/((1 )/(cos x)) = (2sin x)/(cos x) = 2tan x

Finally, we can use the trigonometric identity tan^2 x = sec^2 x - 1 to get:

2tan x =( 2tan^2 x )/( (sec^2 x - 1))

This gives us the following identity:

((tan x + cot x))/((csc x * cos x) ) = sec^2 x

This completes the proof of the trigonometric identity.
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