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2 years ago

Ahmed owns a land of breadth (x+7) units &having an area of 2+ 14x + 49 sq. Units. (i)Find its length.

Mathematics

1 answer:

elena-14-01-66 [18.8K]2 years ago

5 0

Answer:

l = (x+7) units

Step-by-step explanation:

Given that,

Breadth of the rectangle, b = (x+7) units

Area of the rectangle, $A=x^2+14x+49$

The area of a rectangle is given by :

$A=lb\\\\l=\dfrac{A}{b}\\\\l=\dfrac{(x^2+14x+49)}{(x+7)}\\\\l=\dfrac{(x+7)^2}{(x+7)}\\\\l=(x+7)\ \text{units}$

So, the length of the land is (x+7) units.

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Solve and i give brainliest

Temka [501]

Answer:

0.25

Step-by-step explanation:

-9/6 divided by 3/-2 = 0.25

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2 years ago

Read 2 more answers

Leftover Expression

Mumz [18]

The second leftover expression is not o(a+b). It is 6(a + b). I have attached the correct question to depict that.

Answer:

The equivalent expressions are;

8a + 2 and 6a + 6b

Step-by-step explanation:

The two leftover expressions are given as;

2(4x + 1) and 6(a + b)

In algebra, equivalent expressions are simply those expressions which when simplified, give the same resulting expression as the initial one.

Thus simply means expanding or collecting like times to make it clearer.

Now, in our question, like terms have already been collected. This means that to find an equivalent expression, we will just expand the bracket.

Thus;

2(4x + 1) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;

8x + 2

Similarly,

6(a + b) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;

6a + 6b

Thus;

The equivalent expressions are;

8a + 2 and 6a + 6b

4 0

3 years ago

Write the equation of a line that is parallel to the line 2x - 3y = 5 and passes through the point (2, -1)

jek_recluse [69]

Answer:

The equation of the line is 2x - 3y = 7 ⇒ answer A

Step-by-step explanation:

* Lets revise the relation between the parallel lines

- If two lines are equal then their slopes are equal

- We can make an equation of a line by using its slope and

a point on the line

- If the slope of the line is m and passing through the point (x1 , y1),

then we can use this form [y - y1]/[x - x1] = m to find the equation

* Lets solve the problem

- The line is parallel to the line 2x - 3y = 5

∴ the slope of the line = the slope of the line 2x - 3y = 5

- Rearrange the terms of the equation to be in the form

y = mx + c to find the slope of it

∵ 2x - 3y = 5 ⇒ subtract 2x from both sides

∴ -3y = 5 - 2x ⇒ divide two sides by -3

∴ y = 5/-3 - 2x/-3 ⇒ y = 2/3 x - 5/3

∴ The slop of the line is 2/3

∵ The line passes through point (2 , -1)

* Lets use the rule to find the equation of the line

∵ y - (-1)/x - 2 = 2/3

∴ y + 1/x - 2 = 2/3 ⇒ by using cross multiplication

∴ 3(y + 1) = 2(x - 2) ⇒ open the brackets

∴ 3y + 3 = 2x - 4 ⇒ put x an d y in one side

∴ 2x - 3y = 3 + 4

∴ 2x - 3y = 7

* The equation of the line is 2x - 3y = 7

8 0

3 years ago

one-third of the people from country A claim that they are from country B, and the rest admit they are from country A. One-fourt

In-s [12.5K]

Answer: 3 : 2

Step-by-step explanation:

Let A represents the total population of country A and B represents the total population of country B.

According to the question,

$\text{The population of country A that admit they are from B} = \frac{1}{3}\text{ of }A$

⇒ $\text{ The population of A that admit they are from country A }= A - \frac{1}{3} \text{ of } A$

$= \frac{3-1}{3} A$

$= \frac{2}{3} A$

$\text{The population of country B that admit they are from A} = \frac{1}{4}\text{ of }B$

⇒ $\text{ The total population that claims that they are from A }= \frac{2}{3} A +\frac{1}{4} B$

But, Again according to the question,

The total population that claims that they are from A = one half of the total population of A and B.

⇒ $\frac{2}{3} A + \frac{1}{4} B= \frac{1}{2}(A+B)$

⇒ $\frac{2}{3} A + \frac{1}{4} B= \frac{1}{2}A+\frac{1}{2}B$

⇒ $\frac{2}{3} A - \frac{1}{2}A= \frac{1}{2}B-\frac{1}{4} B$

⇒ $\frac{4}{6} A - \frac{3}{6}A= \frac{2}{4}B-\frac{1}{4} B$

⇒ $\frac{1}{6} A = \frac{1}{4} B$

⇒ $A =\frac{6}{4}B$

⇒ $\frac{A}{B} =\frac{3}{2}$

8 0

3 years ago

What are the roots to<br> the equation<br> 2x² + 6x-1=0?

lys-0071 [83]

Answer:

The roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = $\frac{ - b}{a}$

And products of roots = $\frac{c}{a}$

So, m + n = $\frac{ - 6}{2}$ = - 3

And m × n = $\frac{ - 1}{2}$

Or, (m - n)² = (m + n)² - 4mn

Or, (m - n)² = (-3)² - 4 ( $\frac{ - 1}{2}$ )

Or, (m - n)² = 9 + 2 = 11

I.e m - n = $\sqrt{11}$

Again m + n = - 3 And m - n = $\sqrt{11}$

Solving this two equation

(m + n) + ( m - n) = - 3 + $\sqrt{11}$

I.e 2 m = - 3 + $\sqrt{11}$

Or, m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$

Similarly n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Hence the roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$ Answer

6 0

3 years ago