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irakobra [83]
3 years ago
11

2(x-5)=3(x+7) Show the process, please.

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
5 0

Answer:

-31

Step-by-step explanation:

(2×x) - (2×5) = (3×x) + (3×7)

2x-10 = 3x+21

2x-3x = 21+10

-x = 31

therefore, x = -31

Natalija [7]3 years ago
4 0

Answer:

x=11

Step-by-step explanation:

2(x+5)=3(x+7)

2x+10=3x+21

3x-2x=21-10

x=11

Hope this helps! :)

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Answer:

2

Step-by-step explanation:

1+1=2

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I need help with #1 of this problem. It has writings on it because I just looked up the answer because I’m confused but I want t
belka [17]

In the figure below

1) Using the theorem of similar triangles (ΔBXY and ΔBAC),

\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}

Where

\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}

Thus,

\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}

thus, BC = 7.5

2) BX = 9, BA = 15, BY = 15, YC = y

In the above diagram,

\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}

Thus, from the theorem of similar triangles,

\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}

solving for y, we have

\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}

thus, YC = 10.

4 0
1 year ago
Find the volume of a regular hexahedron if one of the diagonals of its faces is 8 root 2 inches.
Naily [24]
The answer is 114sqrt{6} in³

A regular hexahedron is actually a cube. 

Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²

Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²

D² = f² + a²
f² = 2a²

D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3

Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3

V = a³ = (D/√3)³

We have:
D = 8√2 in

V= (\frac{D}{ \sqrt{3} } )^{3} =( \frac{8 \sqrt{2} }{ \sqrt{3} } )^{3}=( \frac{8 \sqrt{2} * \sqrt{3} }{ \sqrt{3}* \sqrt{3}} )^{3} =( \frac{8 \sqrt{2*3} }{ \sqrt{3*3}} )^{3} =( \frac{8 \sqrt{6} }{ \sqrt{3^{2} }} )^{3} =( \frac{8 \sqrt{6} }{ 3} )^{3} = \\ \\ = \frac{ 8^{3} *( \sqrt{6} )^{3}}{3^{3}} = \frac{512* \sqrt{6 ^{2} }* \sqrt{6} }{27} = \frac{512*6* \sqrt{6} }{27} = 114sqrt{6}
8 0
3 years ago
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ArbitrLikvidat [17]

Answer:

I hope This will Help u.. Plz mark me as Brilliant please

Step-by-step explanation:

Before you get started, take this readiness quiz.

Simplify: 

If you missed this problem, review (Figure).

Solve Equations with Constants on Both Sides

In all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we will learn to solve equations in which the variable terms, or constant terms, or both are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the “variable side”, and the other side of the equation to be the “constant side.” Then, we will use the Subtraction and Addition Properties of Equality to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that began with variables and constants on both sides into the form  We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Solve: 

Solution

In this equation, the variable is found only on the left side. It makes sense to call the left side the “variable” side. Therefore, the right side will be the “constant” side. We will write the labels above the equation to help us remember what goes where.



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Use the Subtraction Property of Equality.Simplify.Now all the variables are on the left and the constant on the right.

The equation looks like those you learned to solve earlier.Use the Division Property of Equality.Simplify.Check:Let .

Solve: 



Solve: 



Solve: 

Solution

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4 0
2 years ago
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