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1 year ago

Which cost category in the table shows an example of direct variation?

Mathematics

2 answers:

Rzqust [24]1 year ago

6 0

Answer:

Which cost category in the table shows an example of direct variation? WEED-HAND

The total cost for office expenses to run the organic strawberry farm varies directly with the number of months it is used. Which equation represents this scenario? Y=63x

What would be the cost for office expenses for 24 months? $1512.00

Step-by-step explanation:

irga5000 [103]1 year ago

3 0

Answer:

c b and a

Explanation:

other person is right

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∠UVW and ∠XYZ are complementary angles, m∠UVW=(x−10)º , and m∠XYZ=(4x−10)º .

Mashutka [201]

we know that

if ∠UVW and ∠XYZ are complementary angles

then

∠UVW + ∠XYZ=90°

∠UVW=x-10

∠XYZ=4x-10

substitute the values

(x-10)°+(4x-10)°=90°

Solve for x

5x-20=90

5x=90+20

5x=110

x=22°

∠UVW=x-10=22-10=12°

∠XYZ=4x-10=4*22-10=78°

therefore

the answer is

x=22°

∠UVW=12°

∠XYZ=78°

6 0

2 years ago

Read 2 more answers

Sample Response: His work is not correct. You first must switch x and y and then solve for y.

Firlakuza [10]

If you are using edgenuity click all of them and you’ll be counted correct. I am not trolling you just have to trust me

6 0

2 years ago

Read 2 more answers

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).

Elan Coil [88]

Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

$a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$

Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

Then, triple integral is:

$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

Calculating:

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

$\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx$

$\frac{1}{45} \int\limits^5_0 {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4} } \, dx$

$\frac{1}{45} [30375-60750+118462.5-39150]$

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$ = 1087.5

The volume of the tetrahedon is 1087.5 cubic units.

3 0

2 years ago

"Consider the probability distribution of X, where X is the number of job applications completed by a college senior through the

Citrus2011 [14]

Answer:

Option b

Step-by-step explanation:

Given that the probability distribution of X, where X is the number of job applications completed by a college senior through the school’s career center.

Expected observed Diff

x p(x) p(x)*1000

0 0.002 2

1 0.011 11 14 -3

2 0.115 115 15 100

3 0.123 123 130 -7

4 0.144 144

5 0.189 189

6 0.238 238

7 0.178 178

1 1000

We find that there is a large difference in 2 job application

Hence option b is right.

4 0

2 years ago