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2 years ago

6

A travel agent receives a 9% commission on the base fair of a customer- that is the fair before sales tax is added if the total

fare charged to a customer comes to $784.59 including 5.5% sales tax, how much commission should the agent receive?

1 answer:

Helen [10]2 years ago

3 0

Answer:

98 he should receive yodjct7

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How do i factorise 8x squared plus 4xy ???

andreev551 [17]

8x^2 + 4xy = 4x (2x + y)

hope that helps

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2 years ago

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Help me pls I don’t know how to do this I’ll give brainliest pls help

Natalka [10]

Answer:

3/2

Step-by-step explanation:

First, you find the mean (((4 + 5 + 6 + 1)/4) = 4)

Find the absolute difference between each data value and the mean: 0, 1, 2, 3

Add the differences (in the previous step) and divide by number of terms: (0+1+2+3)/4 = 6/4 = 3/2

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2 years ago

Find the first partial derivatives of the function f(x,y,z)=4xsin(y−z)

Amanda [17]

Answer:

$f_x(x,y,z)=4\sin (y-z)$

$f_x(x,y,z)=4x\cos (y-z)$

$f_z(x,y,z)=-4x\cos (y-z)$

Step-by-step explanation:

The given function is

$f(x,y,z)=4x\sin (y-z)$

We need to find first partial derivatives of the function.

Differentiate partially w.r.t. x and y, z are constants.

$f_x(x,y,z)=4(1)\sin (y-z)$

$f_x(x,y,z)=4\sin (y-z)$

Differentiate partially w.r.t. y and x, z are constants.

$f_y(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial y}(y-z)$

$f_y(x,y,z)=4x\cos (y-z)$

Differentiate partially w.r.t. z and x, y are constants.

$f_z(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial z}(y-z)$

$f_z(x,y,z)=4x\cos (y-z)(-1)$

$f_z(x,y,z)=-4x\cos (y-z)$

Therefore, the first partial derivatives of the function are $f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z)$ .

4 0

3 years ago

The ratio of the length to the width of a picture is 5:2. The picture is enlarged so that its length is 15 inches now. What is t

Maru [420]

The answer is B.
The ratio is multiplied by three so 5 times 3 is 15 and 2 times 3 is 6.

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3 years ago

Solve the equation for principal values of x. Express solutions in degrees.

Hitman42 [59]

Option C:

x = 90°

Solution:

Given equation:

$\sin x=1+\cos ^{2} x$

<u>To find the degree:</u>

$\sin x=1+\cos ^{2} x$

Subtract 1 + cos²x from both sides.

$\sin x-1-\cos ^{2} x=0$

Using the trigonometric identity: $\cos ^{2}(x)=1-\sin ^{2}(x)$

$\sin x-1-\left(1-\sin ^{2}x\right)=0$

$\sin x-1-1+\sin ^{2}x=0$

$\sin x-2+\sin ^{2}x=0$

$\sin ^{2}x+\sin x-2=0$

Let sin x = u

$u^2+u-2=0$

Factor the quadratic equation.

$(u+2)(u-1)=0$

u + 2 = 0, u – 1 = 0

u = –2, u = 1

That is sin x = –2, sin x = 1

sin x can't be smaller than –1 for real solutions. So ignore sin x = –2.

sin x = 1

The value of sin is 1 for 90°.

x = 90°.

Option C is the correct answer.

3 0

2 years ago

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