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vladimir2022 [97]
3 years ago
6

Prove that in general (x + a) 2 ≠ x2 + a2

Mathematics
1 answer:
qwelly [4]3 years ago
6 0

Answer:

yes they are not equal because (x + a) = 2x + 2a

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Christine is 4 years younger than Mark. 3 years from now, she will be two-thirds as old as Mark find their present ages.Complete
icang [17]

Answer:Mark Is 9 years old; Christine is 5 years old

Step-by-step explanation:

Present age

LET  mark age be x

Christine be x-4 yrs

In 3 yrs Time

Mark will be x+3 yrs old

Christine  will be  (x -4 +3) = x-1 years old

Also, In 3 years time

Christine  = 2/3 Mark

such that

  x-1 = 2/3(x+3)

  x -1= 2x+6 / 3

    3x  -3= 2x+6

      3X--2X = 6+3

         =X=9

Mark Is 9 years old

Christine is   x -4 = 9-4=5 years old

5 0
3 years ago
What is the length of the diagonal, d, of the rectangular prism shown below?
Lena [83]

Answer:

d = 10√10 ≈ 31.62

Step-by-step explanation:

d = √l²+w²+h²

d = √5²+8²+2²

d = √25+81+4

d = √110

d = 10*√10

4 0
3 years ago
Read 2 more answers
Find the nonpermissible replacement for a in this expression 7/6a
vredina [299]
The correct answer for the exercise shown above is:

 zero (0)

 The explanation is shown below:

 1. You have the following expression given in the problem above:

 7/6a

 2. By definition, <span>the nonpermissible replacement for a is the value that make the denominator equal to zero.

 3. Keeping this on mind, you have:

 7/6a=7/6(0)=7/0

 Therefore, as you can see, the answer is: zero (a=0).


</span>
6 0
3 years ago
Equilateral triangle ABC has a perimeter of 96 millimeters. A perpendicular bisector is drawn from angle A to side Line segment
prisoha [69]

Answer:

<h2>A  16mm</h2>

Step-by-step explanation:

An equilateral triangle is a triangle that has all of its sides equal.

Perimeter of an equilaterial triangle = 3s where;

s is one side of the triangle.

Given the perimeter of ABC = 96mm

Substituting into the formula above to get s;

96 = 3s

s = 96/3

s = 32mm

Hence the length of one side of the triangle is 32mm

If a perpendicular bisector is drawn from angle A to side Line segment B C at point M, then MC will be half of BC and ΔAMC will be a right angled triangle.

Since all the sides of the triangle are equal, hence BC = 32mm. Since MC is the half of BC, then MC = 1/2 of 32mm

MC = 1/2 * 32mm

MC = 16mm

<em>Hence the length of Line segment MC is 16mm</em>

<em></em>

5 0
3 years ago
Read 2 more answers
A certain forest covers an area of
iVinArrow [24]

Answer:

2598 square kilometers

Step-by-step explanation:

Hello

Step 1

year one

using a rule of three is possible to find how much is 8.75 od 4500 km2

Let

if

4500 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\

at the end of the year one, the area will be

4500-393.75=4106.25

this will be the initial area for the year 2.

Step 2

repite the step 1 with area initial =4106.25 km2

4106.25 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\

at the end of the year 2, the area will be

4106-359.29=3746.70

this will be the initial area for the year 3.

Step 3

repite the step 1 with area initial =4106.25 km2

3746.70 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\

at the end of the year 3, the area will be

3746.70-327.83=3419.09

this will be the initial area for the year  4.

Step 4

year four

repite the step 1 with area initial =3419.09 km2

3419.09 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\

at the end of the year 4, the area will be

3419.09-299.173=3119.82

this will be the initial area for the year  5.

Step 5

year five

repite the step 1 with area initial =3119.82 km2

3119.82 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\

at the end of the year 5, the area will be

3119.82-272.99=2846.92

this will be the initial area for the year  6.

Step 6

year six

repite the step 1 with area initial =2846.92km2

2846.92 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\

at the end of the year six, the area will be

2846.92-249.10=2597.82 square kilometers

Have a great day.

8 0
3 years ago
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