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IrinaK [193]
3 years ago
15

Solve the equation

=" \sqrt{x - 6} + 2 = 6" alt=" \sqrt{x - 6} + 2 = 6" align="absmiddle" class="latex-formula">
for the variable.
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Explanation:

First you had to square from both sides of equation form.

(\sqrt{x-6+2})^2=6^2

Then you expanded form.

(\sqrt{x-6+2})^2=x-4

x-4=36

6^2=6*6=36

Solve by equation form.

x-4=36

Add by four from both sides of equation form.

x-4+4=36+4

Simplify.

x=40

Verified solutions.

x=40 \boxed{TRUE}

Final answer: \boxed{x=40}

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

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A rectangle has perimeter 28cm. Its area is 42sq.cm. Determine the dimensions of the rectangle. Include a diagram in your soluti
yawa3891 [41]

The dimensions of the rectangle are: b=9.65 cm and h=4.35 cm or b=4.35 cm and h= 9.65 cm.

<h3>Quadrilaterals</h3>

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle is equal to base x height (A=bh). The perimeter of a geometric figure is the sum of its sides. Thus, for a rectangle, the perimeter is 2b+2h.

<h3>System of Linear Equations</h3>

System of linear equations is the given term math for two or more equations with the same variables. The solution of these equations represents the point at which the lines intersect.

The question gives:

Perimeter=28cm

Area= 42 cm²

If the perimeter is the sum of all sides of the rectangle, you have:

Perimeter= 2b+2h=28

If the area of the rectangle is 42cm², you have:

Area=bh=42 cm²

You can write a system of linear equations.

2b+2h=28 (1)

bh=42 (2)

From equation 2, you have  b=\frac{42}{h}. Then, you can replace it in equation 1.

2*\frac{42}{h}+2h=28 \\ \\ 84+2h^2=28h\\ \\ 2h^2-28h+84=0 \; dividing\; by\; 2\\ \\ h^2-14h+42=0

Now, you should solve the quadratic equation.

h_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:42}}{2\cdot \:1}

h_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{7}}{2\cdot \:1}

h_1=\frac{-\left(-14\right)+2\sqrt{7}}{2\cdot \:1}=7+\sqrt{7}=9.65 cm

h_2=\frac{-\left(-14\right)-2\sqrt{7}}{2\cdot \:1}=7-\sqrt{7}=4.35cm

If h1=9.65 cm, then b_1=\frac{42}{9.65} =4.35 cm, and if  h2=4.35 cm, then b_1=\frac{42}{4.35} =9.65 cm.

Read more about the quadratic function here:

brainly.com/question/1497716

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8 0
2 years ago
Ms santos seashells into 4 row she puts 6 seashells in each row how many seachells are there altogether
kati45 [8]
4 rows and 6 seashells in each row. 
Row 1: 0 0 0 0 0 0
Row 2: 0 0 0 0 0 0
Row 3: 0 0 0 0 0 0
Row 4: 0 0 0 0 0 0

You add all of them together 6+6+6+6=24 
Or you could do 6x4 and your answer would be 24
So the answer to the question is 24 seashells. 
3 0
3 years ago
Write an expression for the perimeter of the triangle shown in the picture
umka2103 [35]

Answer: x-5+5+5

Step-by-step explanation:

3 0
2 years ago
Find a ⋅ b. a = &lt;5, 2&gt;, b = &lt;4, 5&gt;
cestrela7 [59]

Answer:

  30

Step-by-step explanation:

For a = <xa, ya> and b = <xb, yb>, the dot product is the sum of products ...

  a·b = (xa)(xb) + (ya)(yb)

Substituting the given information, you have ...

  a·b = 5·4 + 2·5 = 20 + 10

  a·b = 30

_____

Some graphing calculators can do such math. There are also dot product calculators available on the Internet. If you have quite a few of these to calculate, you can put the appropriate formula into a spreadsheet.

6 0
3 years ago
hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and
Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
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