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2 years ago

Find the missing dimension of the triangle.

Mathematics

1 answer:

ludmilkaskok [199]2 years ago

3 0

Answer: 16

Step-by-step explanation: 20/2 = 10

10x=160

10x/10=160/10

x=16

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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

Solve (2 - x)^2 < 4/25

Semenov [28]

Answer:

8/5 < x< 12/5

Step-by-step explanation:

(2 - x)^2 < 4/25

Take the square root of each side

sqrt((2 - x)^2) <±sqrt( 4/25)

Make two equations

2-x < 2/5 2-x > -2/5

Subtract 2 from each side

2-x-2 < 2/5 -2 2-x-2 > -2/5-2

-x < 2/5 - 10/5 -x > -2/5 - 10/5

-x < -8/5 -x > -12/5

Multiply by -1, remembering to flip the inequality

x> 8/5 x < 12/5

8/5 < x< 12/5

5 0

3 years ago

Read 2 more answers

Alan drives 20.4 miles in 17 minutes.He passes a sign which gives the speed limit as 60mph.By how much in mph did Alan exceed th

Rzqust [24]

12 miles over the speed limit

7 0

2 years ago

6x−4y=−24<br> 4x−8y=−32 <br><br> What is the ordered pair for both of them?

denis23 [38]

Multiply the first equation by -2 to eliminate a variable so it becomes:
-12x+8y=48
4x-8y=-32
Add them together.
-8x=16
x=-2
Plug it in one equation.
6(2)-4y=-24
12-4y=-24
-4y=-12
y=3
Ordered pair is: (-2,3)

6 0

3 years ago

4 without using an exponent<br> of 1

Allushta [10]

Answer:

what do you mean like 4 with a exponent of 2,3,4,5,6,7,8,9,10?

Step-by-step explanation:

5 0

3 years ago