Please help!!

If the original square had a side length of

irina [24]

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

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

WILL GIVE 100 POINTS AND BRAINLIEST FOR BEST ANSWER

coldgirl [10]

Answer:

Step-by-step explanation:-16x^2 + 24x + 16 = 0.

A. Divide by 8:

-2x^2 + 3x + 2 = 0, A*C = -2*2 = -4 = -1 * 4. Sum = -1 + 4 = 3 = B, -2x^2 + (-x+4x) + 2 = 0,

(-2x^2-x) + (4x+2) = 0,

-x(2x+1) + 2(2x+1) = 0,

(2x+1)(-x+2) = 0, 2x+1 = 0, X = -1/2. -x+2 = 0, X = 2.

X-intercepts: (-1/2,0), (2,0).

B. Since the coefficient of x^2 is negative, the parabola opens downward. Therefore, the vertex is a maximum.

Locate the vertex: h = Xv = -B/2A = -24/-32 = 3/4, Plug 3/4 into the given Eq to find k(Yv). K = -16(3/4)^2 + 16(3/4) + 16 = 19. V(h,k) = V(3/4,19).

C. Choose 3 points above and below the vertex for graphing. Include the points calculated in part A which shows where the graph crosses the x-axis.

4 0

3 years ago

In triangle abc, the perimeter is 36, and ab=ac. line ad is perpendicular to bc at the point d, and triangle abd's perimeter is

Musya8 [376]

Let x=ab=ac, and y=bc, and z=ad.

Since the perimeter of the triangle abc is 36, you have:

Perimeter of abc = 36
ab + ac + bc = 36
x + x + y = 36
(eq. 1) 2x + y = 36

The triangle is isosceles (it has two sides with equal length: ab and ac). The line perpendicular to the third side (bc) from the opposite vertex (a), divides that third side into two equal halves: the point d is the middle point of bc. This is a property of isosceles triangles, which is easily shown by similarity.

Hence, we have that bd = dc = bc/2 = y/2 (remember we called bc = y).

The perimeter of the triangle abd is 30:

Permiter of abd = 30
ab + bd + ad = 30
x + y/2 + z =30
(eq. 2) 2x + y + 2z = 60

So, we have two equations on x, y and z:

(eq.1) 2x + y = 36
(eq.2) 2x + y + 2z = 60

Substitute 2x + y by 36 from (eq.1) in (eq.2):

(eq.2') 36 + 2z = 60

And solve for z:

36 + 2z = 60 => 2z = 60 - 36 => 2z = 24 => z = 12

The measure of ad is 12.

If you prefer a less algebraic reasoning:

- The perimeter of abd is half the perimeter of abc plus the length of ad (since you have "cut" the triangle abc in two halves to obtain the triangle abd).

- Then, ad is the difference between the perimeter of abd and half the perimeter of abc:

ad = 30 - (36/2) = 30 - 18 = 12

4 0

3 years ago

Read 2 more answers

PLEASE SOMEONE HELP ME

mylen [45]

Answer:

.

Step-by-step explanation:

4 0

3 years ago

What is the answer to(1−5q)+2(2.5q+8)

never [62]

Greetings!

Simplify the Expression.
$=(1-5q)+2(2.5q+8)$

Distribute the Parenthesis.
<em>How?</em><span> Multiply the terms inside the Parenthesis by the term outside of the Parenthesis.
</span> $=1*1-1*5q+2*2.5q+2*8$

Combine Like Terms.
$=1-5q+5q+16$

$=17$

The Answer Is:
$\left[\begin{array}{ccc}17\end{array}\right]$

Hope this helps.
-Benjamin

5 0

3 years ago