After school, a student jogged from school to the library. it took the student 35 minutes to get to the library, where he spent 10 minutes browsing and checking out a book. he then walked to the coffee shop, on the same street, in 15 minutes. he spent 40 minutes sitting at a table in the coffee shop and reading the book before heading home. he walked home in 25 minutes and remain there until the next morning.

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Answer 1

Answer:

Answer:

so what are you asking exactly

Step-by-step explanation:

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Find the number of distinct triangles with a=11, b=17, and A=23°A. 0
B. 1
C. 2
D. cannot be determined

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The answer would be C, 2

HELP PLEZ TRIGONOMETRY!

Answers

$(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha$

Solution:

Given that we have to simplify:

$(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)$ ---- eqn 1

We know that,

$sin^2 x = 1 - cos^2 x$

Substitute the above identity in eqn 1

$(2\left(1-\cos ^(2) \alpha\right)-1)/(\sin \alpha+\cos \alpha)$

Simplify the above expression

$(2-2 \cos ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)$

$(1-2 \cos ^(2) \alpha)/(\sin \alpha+\cos \alpha)$ ------- eqn 2

By the trignometric identity,

$(sin x + cos x)(sin x - cos x) = 1-2cos^2 x$

Substitute the above identity in eqn 2

$((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)$

Cancel the common factors in numerator and denominator

$((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)=\sin \alpha-\cos \alpha$

Thus the simplified expression is:

$(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha$

Suppose f(x,y)=xy, P=(−4,−4) and v=2i+3j. A. Find the gradient of f. ∇f= i+ j Note: Your answers should be expressions of x and y; e.g. "3x - 4y" B. Find the gradient of f at the point P. (∇f)(P)= i+ j Note: Your answers should be numbers C. Find the directional derivative of f at P in the direction of v. Duf= Note: Your answer should be a number D. Find the maximum rate of change of f at P. Note: Your answer should be a number E. Find the (unit) direction vector in which the maximum rate of change occurs at P. u= i+ j Note: Your answers should be numbers

Answers

Answers:

Gradient of f: $\nabla f = y\hat{i} + x\hat{j}$

Gradient of f at point p: $\nabla f = -4\hat{i} -4\hat{j}$

Directional derivative of f and P in direction of v: $\nabla f(P)v = -20\n$

The maximum rate of change of f at P: $| \nabla f(P)| = 4√(2)$

The (unit) direction vector in which the maximum rate of change occurs at P is: $v = -(1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}$

Step by step solutions:

Given that:

$f(x,y) = xy$

$P = (-4,4)\n$

$v = 2i + 3j$

A: Gradient of f

$\nabla f = ((\partial f)/(\partial x), (\partial f)/(\partial y)) = (y,x) = y\hat{i} + x\hat{j}$

B: Gradient of f at point P:

Just put the coordinates of p in above formula:

$\nabla f = -4\hat{i} -4\hat{j}$

C: The directional derivative of f and P in direction of v:

The directional derivative is found by dot product of $\nabla f(P) \: \rm and \: \rm v$ :

$\nabla f(P)v = [-4,4][2,3]^T = -20\n$

D: The maximum rate of change of f at P is calculated by evaluating the magnitude of gradient vector at P:

$| \nabla f(P)| = √((-4)^2 + (-4)^2) = 4√(2)$

E: The (unit) direction vector in which the maximum rate of change occurs at P is:

$v = ((-4)/(4√(2)), (-4)/(4√(2))) = -(1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}$

That vector v is the needed unit vector in this case.

we divided by $4√(2)$ to make that vector as of unit length.

Learn more about vectors here:

brainly.com/question/12969462

Answer:

a) The gradient of a function is the vector of partial derivatives. Then

$\nabla f=((\partial f)/(\partial x), (\partial f)/(\partial y))=(y,x)=y\hat{i} + x\hat{j}$

b) It's enough evaluate P in the gradient.

$\nabla f(P)=(-4,-4)=-4\hat{i} - 4 \hat{j}$

c) The directional derivative of f at P in direction of V is the dot produtc of $\nabla f(P)$ and v.

$\nabla f(P) v=(-4,-4)\left[\begin{array}{ccc}2\n3\end{array}\right] =(-4)2+(-4)3=-20$

d) The maximum rate of change of f at P is the magnitude of the gradient vector at P.

$||\nabla f(P)||=√((-4)^2+(-4)^2)=√(32)=4√(2)$

e) The maximum rate of change occurs in the direction of the gradient. Then

$v=(1)/(4√(2))(-4,-4)=((-1)/(√(2)),(-1)/(√(2)))= (-1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}$

is the direction vector in which the maximum rate of change occurs at P.

The graph of f(x) = x2 was transformed to create the graph of g(x) = (x − 7.5)2. Which of these describes this transformation?A.
A horizontal shift to the right 7.5 units

B.
A horizontal shift to the left 7.5 units

C.
A vertical shift down 56.25 units

D.
A vertical shift up 56.25 units

Answers

Answer:

A. A horizontal shift to the right 7.5 units

Step-by-step explanation:

Replacing x with x-7.5 shifts the graph 7.5 units to the right.

In general, g(x) = f(x-h)+k will shift h units right and k units up. In this problem there is no vertical shift.

A candy distributor needs to mix a 30% fat-content chocolate with a 50% fat-content chocolate to create 200 kilograms of a 46% fat-content chocolate. How many kilograms of each kind of chocolate must they use?

Answers

Step-by-step explanation:

If x is the kilograms of 30% chocolate, and y is the kilograms of 50% chocolate, then:

x + y = 200

0.30x + 0.50y = 0.46(200)

Solving the system of equations with substitution:

0.30x + 0.50(200 − x) = 0.46(200)

0.30x + 100 − 0.50x = 92

8 = 0.20x

x = 40

y = 200 − x

y = 160

The distributor needs 40 kg of 30% chocolate and 160 kg of 50% chocolate.

Final answer:

To obtain 200 kilograms of a 46% fat-content chocolate, the candy distributor needs to mix 40 kilograms of a 30% fat-content chocolate and 160 kilograms of a 50% fat-content chocolate.

Explanation:

This problem can be solved using a basic mixture problem method. Let's name the amount of the 30% fat-content chocolate as 'x' and the amount of the 50% fat-content chocolate as 'y'. The total weight of the resulting chocolate is provided in the problem, 200 kilograms, therefore we know that x + y = 200.

The total fat in the chocolates should be 46% of 200kg, or 92kg. This gives us another equation based on the fat content, 0.3x + 0.5y = 92.

Solving these two equations linearly, we find the values of x and y. The amount of 30% fat content chocolate (x) is 40 kilograms and the amount of 50% fat-content chocolate (y) is 160 kilograms.

Learn more about Mixture Problem here:

brainly.com/question/33435099

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What is the initial value of the exponential function shown on the graph? a. 0, b. 1, c. 2, d. 4

Answers

Answer:

Step-by-step explanation:

4 is the initial value of the exponential function shown on the graph.

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