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2 years ago

Solve this equation: -7x + 12 – 2x = 23 + 13x

Mathematics

1 answer:

Arlecino [84]2 years ago

3 0

Answer:

x= -1.2

Step-by-step explanation:

Simplifying

-7 + 12 + -2x = 23 + 13x

Combine like terms: -7 + 12 = 5

5 + -2x = 23 + 13x

Solving

5 + -2x = 23 + 13x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-13x' to each side of the equation.

5 + -2x + -13x = 23 + 13x + -13x

Combine like terms: -2x + -13x = -15x

5 + -15x = 23 + 13x + -13x

Combine like terms: 13x + -13x = 0

5 + -15x = 23 + 0

5 + -15x = 23

Add '-5' to each side of the equation.

5 + -5 + -15x = 23 + -5

Combine like terms: 5 + -5 = 0

0 + -15x = 23 + -5

-15x = 23 + -5

Combine like terms: 23 + -5 = 18

-15x = 18

Divide each side by '-15'.

x = -1.2

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2 years ago

The square shown has a perimeter of 32 units. The square is rotated about line k.

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A cylinder with a circumference of about 50 units is formed.

Step-by-step explanation:

Step 1:

The perimeter of the square is 32 units. The perimeter of a square is given by 4 times its side length.

$(4) sidelength = 32, sidelength = \frac{32}{4} = 8.$

So the side length of the square is 8 units.

If a square is rotated a cylinder will be formed. So the options with a cone are wrong as a cone is formed when a triangle is rotated.

Step 2:

To calculate the circumference of the cylinder, we multiply 2π with the radius. The radius is the same as the side length of the square. So r is 8 units.

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So the circumference of the cylinder is approximately equal to 50 units so the answer is the fourth option, a cylinder with a circumference of about 50 units.

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2 years ago

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If you have 1/8 in every loaf and 1 teaspoon of nutmeg is 8/8 and you have 4 of those think how much is 8/8 x 4.

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Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22

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Answer:

$b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$

Step-by-step explanation:

Consider the given matrix

$A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]$

Let matrix B is

$B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

It is given that

$A=B$

$\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

On comparing corresponding elements of both matrices, we get

$b_{11}=9,b_{12}=-2,b_{13}=3$

$b_{21}=2,b_{22}=17,b_{23}=0$

$b_{31}=3,b_{32}=22,b_{33}=8$

Therefore, the required values are $b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$ .

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With that being said, your function could be something like this:

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