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3 years ago

6

You purchase 26 “parking hours” that you can use over the next month to park your food truck at the fair. Weekday hours costs $2

/hour and weekend hours cost $10/hour. You spent a total of $220. How many weekday hours did you purchase?

1 answer:

Oksanka [162]3 years ago

6 0

Answer: 5

Step-by-step explanation:

weekday hours x*2

weekend hours y*10

x+y=26

2x+10y=220

----------------------

x=26-y

2(26-y)+10y=220

52-2y+10y=220

8y=220-52

8y=168

y=168/8

y=21

x=26-21=5

x=5

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Harlamova29_29 [7]

Answer:

2/8 or 1/4

Step-by-step explanation:

2/8 or 1/4 because the spots on the spinner are blue out of 8 is 2/8 or 1/4

3 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Find the value of cos x° + sin yº

MA_775_DIABLO [31]

Answer:

1.84

Step-by-step explanation:

The given arc is a part of a unit circle centered at origin.

The x-coordinate on the unit circle represent the cosine value and y-coordinate on the unit circle represent sine value.

So, for angle equal to $x$ °, the x coordinate represents $cos (x)$ °.

So, $cos x$ ° = 0.86

Similarly, the y-coordinate of angle y represent $\sin y$ °.

So, $\sin y$ ° = 0.98

Therefore, the value of $cos (x)$ ° + $\sin y$ ° = 0.86 + 0.98 = 1.84.

4 0

3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$

Since 9 is greater than 0, therefore, function is increasing on interval $(-\infty, -1) \text{ or } x$ .

Now we will check 0 for 2nd interval.

$f'(0)=3(0)^{2}-3=0-3=-3$

Since -3 is less than 0, therefore, function is decreasing on interval $(-1,1) \text{ or } -1$ .

We will check 2 from our 3rd interval.

$f'(2)=3(2)^{2}-3=3(4)-3=12-3=9$

Since 9 is greater than 0, therefore, function is increasing on interval $(1,\infty) \text{ or } x>1$ .

6 0

4 years ago

Can someone help me with this question? I’ll mark you brainliest!!!

Mariulka [41]

Answer:

A. P' = (9, -2)

B. P" = (1, -4)

Step-by-step explanation:

To make it easier to understand, make point C the origin (0 , 0). From that, you will notice point P is 3 units right and 5 units up of point C, meaning the coordinates of point P is (3, 5). A 90° rotation clockwise on point P (3, 5) is (5, -3). Since point C is 4 units right and 1 unit up from the actual origin, we will add that to (5,-3); (5 + 4, -3 + 1) = (9, -2). Therefore, P' = (9, -2).

Knowing the coordinates of P' allows us to figure out the coordinates of P''. Once again, we'll make point C the origin, meaning P' would be at (5, -3). Like we did before, we will perform a 90° rotation clockwise of (5, -3) to get (-3, -5). Lastly, add back the appropriate amount of units to (-3, -5) to make point C the center of rotation in this example; (-3 + 4, -5 + 1) = (1, -4). Therefore, P" = (1, -4).

Please note that this is only my method of solving translations. Please consult more official sources if you want to learn more. Other than that, I hope this helps!

7 0

3 years ago