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Mrac [35]
3 years ago
11

What do you call the absolute values of 6 and -6

Mathematics
1 answer:
coldgirl [10]3 years ago
3 0

I believe they are called opposites.

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I NEED HELP.. :(<br><br> Someone please help me with this edge unity question
nasty-shy [4]

Answer:

I think its a,c,d,f, and g

7 0
3 years ago
Wich calculation will ALWAYS have a result greater than 1​
klemol [59]

Step-by-step explanation:

number less than 1 B.4/9+ a fraction less than 12 C. 1 3/4 - a fraction less than 3/4 D.7/8 a number less than 1

3 0
3 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
El perímetro de un rectángulo es 58 y su base excede en 10 a su ancho, ¿Cuánto mide la base? AYUDAAAA
kobusy [5.1K]

Answer:

The base is 19.5.

Step-by-step explanation:

The given question is, "The perimeter of a rectangle is 58 and its base exceeds its width by 10, how long is the base?"

Perimeter = 58

Base, l = 10+b

The perimeter of a rectangle is :

P = 2(l+b)

58 = 2(10+b+b)

29 = (10+2b)

29-10 = 2b

19 = 2b

b = 9.5

Base, l = 10 + 9.5

= 19.5

Hence, the base is 19.5.

5 0
3 years ago
Solve 5,4 = n/-0.9
PtichkaEL [24]

Step-by-step explanation:

5.4×-0.9=n

n=-4.86

(-4/5)/(1/3)=x

X=-2 2/5

8 0
3 years ago
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