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3 years ago

Question 5

Mathematics

1 answer:

nadezda [96]3 years ago

4 0

When you distribute you should get
$12 {x}^{2} + 8x - 3x - 2$
Simplify to get
$12 {x}^{2} + 5x - 2$
Answer 2 is correct

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I camper attaches a rope from the top of her tent 4 feet above the ground to give it more support if the Rope is 8 ft long about

iVinArrow [24]

Answer:

b = 6.928

Step-by-step explanation:

Looks like a job for the Pythagorean Theorem.

a^2 + b^2 = c^2

4^2 + b^2 = 8^2

16 + b^2 = 64

b^2 = 48

b = 6.928

Hope it helps!!

6 0

4 years ago

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

PLEASE HELP WITH THIS QUESTION WILL GIVE BRAINLY

Vadim26 [7]

Answer:

Z' = (-5,3)

Step-by-step explanation:

Move two to the (<- left) from -3,5 to get -5, 5

move two down to get -5,3

8 0

3 years ago

Solve all of them please u get 30 points

nika2105 [10]

Step-by-step explanation:

from which text book please

8 0

2 years ago

Read 2 more answers

Daisy is building a rectangular pen for her chickens along one wall in her back yard. She needs to build a fence

frosja888 [35]

Answer:

Daisy has one side of the rectangular pen along the wall, so she needs one width, w, and two lengths, 2ℓ. She has at most 125 feet of fencing materials to build the fence for her pen

1 second/24 frames

so if he started with 0 frames

time s = (1/24) (additional f)

but he already had at least 250 frames

time s = (1/24)( f-250) or maybe less time if he had more than 250

24 s </= f - 250

f >/= 24 s + 250

Step-by-step explanation:

7 0

3 years ago