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mihalych1998 [28]
3 years ago
13

Sqrt of 4a-4 + sqrt of a-1

Mathematics
1 answer:
Monica [59]3 years ago
5 0
\large\begin{array}{l} \textsf{Simplify the expression (I'm goint to call it E)}\\\\ \mathsf{E=\sqrt{4a-4}+\sqrt{a-1}} \end{array}


\large\begin{array}{l} \textsf{So E is a sum of two square roots. Notice that you can factor}\\\textsf{the 4 out on the first one:}\\\\ \mathsf{E=\sqrt{4\cdot a-4\cdot 1}+\sqrt{a-1}}\\\\ \mathsf{E=\sqrt{4\cdot (a-1)}+\sqrt{a-1}}\\\\ \mathsf{E=\sqrt{2^2\cdot (a-1)}+\sqrt{a-1}}\\\\ \mathsf{E=\sqrt{2^2}\cdot \sqrt{a-1}+\sqrt{a-1}}\\\\ \mathsf{E=2\cdot \sqrt{a-1}+\sqrt{a-1}} \end{array}

\large\begin{array}{l} \mathsf{E=2\cdot \sqrt{a-1}+1\cdot \sqrt{a-1}}\\\\ \mathsf{E=(2+1)\cdot \sqrt{a-1}}\\\\ \boxed{\begin{array}{l}\mathsf{E=3\cdot \sqrt{a-1}} \end{array}} \end{array}


If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2179170


\large\textsf{I hope this helps. :-)}


Tags: <em>square root sqrt simplify expression radical irrational algebra</em>

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Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas  in the attached figure N 3, and the trinomial is x^{2} +11x+28

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

Length=(x+4)\ in

width=(x+7)\ in

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas  in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

A1=(x)(x)=x^{2}\ in^2

A2=(4)(x)=4x\ in^2

A3=(x)(7)=7x\ in^2

A4=(4)(7)=28\ in^2

The total area of the second rectangle is the sum of the four areas

A=A1+A2+A3+A4

State the product of (x+4) and (x+7) as a trinomial

(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28

Part c) If the original square had a side length of  x = 2 inches, then what is the area of the  second rectangle?

we know that

The area of the second rectangle is equal to

A=A1+A2+A3+A4

For x=2 in

substitute the value of x in the area of each portion

A1=(2)(2)=4\ in^2

A2=(4)(2)=8\ in^2

A3=(2)(7)=14\ in^2

A4=(4)(7)=28\ in^2

A=4+8+14+28

A=54\ in^2

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

A(x)=x^{2} +11x+28

For x=2 in

substitute and solve for A(x)

A(2)=2^{2} +11(2)+28

A(2)=4 +22+28

A(2)=54\ in^2 ----> verified

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The trinomial represent the total area of the second rectangle

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