All categories

2 years ago

The following figures are not drawn to scale but AB and CD are straight lines. Find x.

Mathematics

2 answers:

leonid [27]2 years ago

7 0

Answer:

x = 30

Step-by-step explanation:

2x + 90° + x = 180°

3x + 90° = 180°

3x = 90°

x = 90/3

x = 30

PSYCHO15rus [73]2 years ago

4 0

Answer:

Step-by-step explanation:

90° + x + x+ x = 180 {Straight line angles}

90 + 3x = 180

Subtract 90from both sides

3x = 180- 90

3x = 90

Divide both sides by 3

x = 90/3

x = 30

You might be interested in

Angles and polygons

Andre45 [30]

Answer:

∠1 = 66°

Step-by-step explanation:

∠1 + ∠3 = ∠4 (exterior angle theorem)

∠1 + 57° = 123°

∠1 = 66°

8 0

3 years ago

A boy has 3 red , 2 yellow and 3 green marbles. In how many ways can the boy arrange the marbles in a line if: a) Marbles of the

Ilya [14]

Answer:

The total number of different arrangements is 560.

Step-by-step explanation:

A multiset is a collection of objects, just like a set, but can contain an object more than once.

The multiplicity of a particular type of object is the number of times objects of that type appear in a multiset.

Permutations of Multisets Theorem.

The number of ordered n-tuples (or permutations with repetition) on a collection or multiset of $n$ objects, where there are $k$ kinds of objects and object kind 1 occurs with multiplicity $n_1$ , object kind 2 occurs with multiplicity $n_2$ , ... , and object kind $k$ occurs with multiplicity $n_k$ is:

$\begin{equation*}\frac{n!}{n_1!*n_2!*\dots * n_k!}\end{equation*}$

We know that a boy has 3 red, 2 yellow and 3 green marbles. In this case we have n = 8.

If marbles of the same color are indistinguishable, then the total number of different arrangements is

${8 \choose 3, 2, 3} = \frac{8 !}{3 ! 2 ! 3 !} = \frac{8\cdot \:7\cdot \:6\cdot \:5\cdot \:4}{2!\cdot \:3!}=\frac{6720}{2!\cdot \:3!}=\frac{6720}{12}=560$

8 0

3 years ago

Two angles in a triangle equal 20 and 130. What is the measure of the third angle?

mezya [45]

C because they all have to add up to 180 and 20+130=150 and 180-150=30

4 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Cleft.%20%5Cbegin%7Bcases%7D%20%7B%208%20x%20%2B%202%20y%20%3D%2046%20%7D%20%5C%5C%20%7B%207

Brut [27]

$\huge \boxed{\mathfrak{Question} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

First, write both the equations in its standard form.

$8x+2y=46\\ 7x+3y=47$

Now, write the equations in form of matrix.

$\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)$

Then, multiply the equation towards the left by using the inverse of matrix $\left(\begin{matrix}8&2\\7&3\end{matrix}\right)$

$\sf \: inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

The product of the matrix & its inverse will be the identity matrix.

$\sf\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

Now, multiply the matrices that lie on the left-hand side of the equal sign.

$\sf\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

For the 2 × 2 matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ , the inverse matrix is ⇨ $\left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right)$ .

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Do the calculations.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Multiply the matrices.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)$