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Anarel [89]
3 years ago
12

I’m really confused How??? Answer???

Mathematics
2 answers:
Oduvanchick [21]3 years ago
8 0

Answer:

x = 3

Step-by-step explanation:

6x+ 5 = 3x + 14

Subtract 3x from both sides

6x+ 5 -3x = 3x + 14 - 3x

Simplifying

3x + 5 = 14

Subtract 5 from both sides

3x + 5 - 5 = 14 - 5

Simplifying

3x = 9

Divide both sides by 3

3x/3  = 9/3

Simplifying

x = 3

dlinn [17]3 years ago
6 0

Answer:

x=3

Step-by-step explanation:

First, you need to take 3x from both sides.

6x+5-3x=3x+14-3x (3x+5=14).

Next, subtract 5 from both sides.

3x=9

Finally, divide each side by 3.

3x÷3=9÷3 (x=3).

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Find the interquartile range of the data represented by this box plot.
Eduardwww [97]

Answer:

B) 30

Step-by-step explanation:

The interquartile range in 100 to 130.

130-100=30

7 0
2 years ago
Read 2 more answers
Find all of the eigenvalues λ of the matrix A. (Hint: Use the method of Example 4.5 of finding the solutions to the equation 0 =
Svetradugi [14.3K]

Answer:

\lambda=8,\ \lambda=-5

Step-by-step explanation:

<u>Eigenvalues of a Matrix</u>

Given a matrix A, the eigenvalues of A, called \lambda are scalars who comply with the relation:

det(A-\lambda I)=0

Where I is the identity matrix

I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

The matrix is given as

A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]

Set up the equation to solve

det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0

Expanding the determinant

det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0

(3-\lambda)(-\lambda)-40=0

Operating Rearranging

\lambda^2-3\lambda-40=0

Factoring

(\lambda-8)(\lambda+5)=0

Solving, we have the eigenvalues

\boxed{\lambda=8,\ \lambda=-5}

8 0
3 years ago
onyango wishes to fence a rectangular research plot using 100 m of wire .one end of the plot has a wall already erected .calcula
earnstyle [38]

Answer:

1250 m²

Step-by-step explanation:

Let x and y denote the sides of the rectangular research plot.

Thus, area is;

A = xy

Now, we are told that end of the plot already has an erected wall. This means we are left with 3 sides to work with.

Thus, if y is the erected wall, and we are using 100m wire for the remaining sides, it means;

2x + y = 100

Thus, y = 100 - 2x

Since A = xy

We have; A = x(100 - 2x)

A = 100x - 2x²

At maximum area, dA/dx = 0.thus;

dA/dx = 100 - 4x

-4x + 100 = 0

4x = 100

x = 100/4

x = 25

Let's confirm if it is maximum from d²A/dx²

d²A/dx² = -4. This is less than 0 and thus it's maximum.

Let's plug in 25 for x in the area equation;

A_max = 25(100 - 2(25))

A_max = 1250 m²

5 0
3 years ago
What is mathematically incorrect about the claim on the product? (Note the 90" is crossed out , so the jar contains 115 pads ins
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8 0
2 years ago
Read 2 more answers
I only need the answer to question B
nalin [4]

Answer:

x=5

y=1

Step-by-step explanation:

The substitution method is considered easier because of its simplification.

2x-3y=7 (1)

x+2y=7 (2)

considering equation (2),

x = 7 - 2y (3)

using the value of "x" in (1)

2 (7-2y) - 3y=7

14 - 4y- 3y = 7

14 - 7y = 7

7y = 14 - 7

7y = 7

y = 1 (4)

substituting this value of "y" in (3)

x = 7- 2(1)

x = 5

3 0
3 years ago
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