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4 years ago

Is a rectangle a rhombus

Mathematics

2 answers:

jek_recluse [69]4 years ago

4 0

Answer:

Yes!

Step-by-step explanation:

That’s because a square is a parallelogram with four congruent sides and four right angles.

Travka [436]4 years ago

3 0

Yes a rectangle is a rhombus

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Which of the following is the result of expanding

andrey2020 [161]

The given expression is 2x-3 .

So when x =1, we will get 2(1)-3 = 2-3 =-1

When x=2, we will get, 2(2) -3 = 4-3 =1

When x=3, we will get 2(3)-3 = 6-3=3

When x=4, we will get 2(4)-3 = 8-3=5

So the required result of the expansion is -1+1+3+5=8

Correct option is the second option .

7 0

4 years ago

Read 2 more answers

Solve each equation x-6=8

jok3333 [9.3K]

Equation is
x-6=8
x=8+6
x=14
X equal to 14

8 0

3 years ago

How to find the perimeter of the triangle

Sliva [168]

Answer:

A. 26.2

Step-by-step explanation:

To find the perimeter of the triangle, you have to find the distances of all three lines and add them up.

Line AB

Let's start off by finding the distance of line AB.

We will use the formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point B is (4,-3).

To substitute the values, it will get to $d = \sqrt{(4--2)^{2}+(-3-5)^{2}}$ which in other words is $d = \sqrt{(4+2)^{2}+(-3-5)^{2}}$ .

Now we have to solve the parenthesis to get $d = \sqrt{(6)^{2}+(-8)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{36+64}$ .

Now we have to simplify the square root to $d = \sqrt{100}$ . In other words, that is $d = 10$ .

Line AB = 10

Line BC

Now let's find the distance of line BC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point B is (4,-3) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0-4)^{2}+(-6--3)^{2}}$ which in other words is $d = \sqrt{(0-4)^{2}+(-6+3)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(-4)^{2}+(-3)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{16+9}$ .

Now we have to simplify the square root to $d = \sqrt{25}$ . In other words, that is $d = 5$ .

Line BC = 5.

Line AC

Now let's fine the distance of line AC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0--2)^{2}+(-6-5)^{2}}$ which in other words is $d = \sqrt{(0+2)^{2}+(-6-5)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(2)^{2}+(-11)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{4+121}$ .

Now we have to simplify the square root to $d = \sqrt{125}$ . In other words, that is $d = 5\sqrt{5}$ . To round that, $d = 11.2$ .

Line AC = 11.2.

Perimeter of Triangle ABC

Perimeter of Triangle ABC = Line AB + Line BC + Line AC.

Perimeter of Triangle ABC = 10 + 5 + 11.2

Perimeter of Triangle ABC = 26.2

Hope this helped! If not, please let me know <3

3 0

3 years ago

The points -5,2 and -5,8 lie on a circle with a radius of 3. Find the center of the circle.

worty [1.4K]

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have two points A(-5, 2) and B(-5, 8). Substitute:

$d=\sqrt{(8-2)^2+(-5-(-5))^2}=\sqrt{6^2+0^2}=\sqrt{6^2}=6$

d is a diameter of a circle because d = 2r = 2(3).

The center of a circle is in midpoint of AB.

The formula of a midpoint:

$M\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$\dfrac{-5+(-5)}{2}=\dfrac{-10}{2}=-5\\\\\dfrac{2+8}{2}=\dfrac{10}{2}=5$

Answer: (-5, 5).

6 0

3 years ago

Sofian's current age is (2x + 3) times Mastura's age. Given that Mastura's current age is (x + 5) years and 6 years ago, the sum

Sliva [168]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

★ Sofian's current age is (2x + 3) times Mastura's age.

★ Mastura's current age is (x + 5) years

★ 6 years ago, the sum of their ages is 44 years.

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

★ The value of x.

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

According to the question,

Present age of Mastura = (x + 5) years

Present age of Sofian = (2x + 3) (x + 5) years, i.e :

$\longrightarrow \tt (2x + 3) (x + 5)$

$\longrightarrow \tt 2 {x}^{2} + 10x + 3x + 15$

$\longrightarrow \tt 2 {x}^{2} + 13x + 15 \: years$

Now,

Mastura's age 6 years ago = (x + 5) - 6

=> x + 5 - 6

=> x - 1

Sofian's age 6 years ago = 2x² + 13x + 15 - 6

= 2x² + 13x + 9

Using the condition provided by the question, 

Mastura's age 6 years ago + Sofian's age 6 years ago = 44

$\longrightarrow \tt (x - 1) + (2 {x}^{2} + 13x + 9) = 44$

$\longrightarrow \tt x - 1 + 2 {x}^{2} + 13x + 9 = 44$

$\longrightarrow \tt 2 {x}^{2} + 14x + 8 = 44$

$\longrightarrow \tt 2 {x}^{2} + 14x + 8 - 44 = 0$

$\longrightarrow \tt 2 {x}^{2} + 14x - 36= 0$

• Taking "2" common.

$\longrightarrow \tt 2 ({x}^{2} + 7x - 18) = 0$

$\longrightarrow \tt {x}^{2} + 7x - 18 = \dfrac{0}{2}$

$\longrightarrow \tt {x}^{2} + 7x - 18= 0$

• Using splitting the middle term

$\longrightarrow \tt {x}^{2} + 9x - 2x - 18 = 0$

$\longrightarrow \tt x(x + 9) - 2(x + 9) = 0$

$\longrightarrow \tt (x + 9) (x - 2) = 0$

either $\tt (x + 9) = 0 \: \: or \: \: (x - 2) = 0$

$\longrightarrow \tt x = - 9 \: \: or \: \: x = 2$

[As the age cannot be negative. So x = - 9 is rejected]

$\therefore \tt \: x = \red{ 2}$

${\large{\textsf{\textbf{\underline{\underline{Verification :}}}}}}$

According to the condition provided by the question, 

• Mastura's age 6 years ago + Sofian's age 6 years ago = 44

$\longrightarrow \tt x - 1 + 2 {x}^{2} + 13x + 9 = 44$

Putting x = 2 we get,

$\longrightarrow \tt 2 - 1 + 2( {2})^{2} + 13(2) + 9 = 44$

$\longrightarrow \tt 1 + 2 \times 4 + 26+ 9 = 44$

$\longrightarrow \tt 1 + 8 + 26+ 9 = 44$

$\longrightarrow \tt 9 + 26+ 9 = 44$

$\longrightarrow \tt 18 + 26 = 44$

$\longrightarrow \tt 44 = 44$

Hence verified.

$\begin{gathered} {\underline{\rule{290pt}{3pt}}} \end{gathered}$

Hope it helps you! :))

6 0

3 years ago