All categories

3 years ago

What is the value of x?Enter your answer in the box.x=

Mathematics

1 answer:

Lina20 [59]3 years ago

3 0

Answer:

x = 8

Step-by-step explanation:

This is an equilateral triangle, so we know that the angles are all equal too.

That means each angle is 60 degrees (b/c 60 * 3 = 180)

7x + 4 = 60

7x = 56

x = 8

You might be interested in

What is equivalent to 4/9(2n-3)

sp2606 [1]

4/9(2n)+4/9(-3)
(4x2/9)n+(4 x -3)/9
(8/9)n -12/9
8/9n-4/3

4 0

3 years ago

"Ten subtracted from the quotient of a number and<br>7 is less than -6.

RideAnS [48]

Written out, this would be (x / 7) - 10 < -6

solved, it would be x < 28

5 0

3 years ago

Find out the number of combinations and the number of permutations for 8 objects taken 6 at a time. Express your answer in exact

umka2103 [35]

Solution:

The permutation formula is expressed as

$\begin{gathered} P^n_r=\frac{n!}{(n-r)!} \\ \end{gathered}$

The combination formula is expressed as

$\begin{gathered} C^n_r=\frac{n!}{(n-r)!r!} \\ \\ \end{gathered}$

where

$\begin{gathered} n\Rightarrow total\text{ number of objects} \\ r\Rightarrow number\text{ of object selected} \end{gathered}$

Given that 6 objects are taken at a time from 8, this implies that

$\begin{gathered} n=8 \\ r=6 \end{gathered}$

Thus,

Number of permuations:

$\begin{gathered} P^8_6=\frac{8!}{(8-6)!} \\ =\frac{8!}{2!}=\frac{8\times7\times6\times5\times4\times3\times2!}{2!} \\ 2!\text{ cancel out, thus we have} \\ \begin{equation*} 8\times7\times6\times5\times4\times3 \end{equation*} \\ \Rightarrow P_6^8=20160 \end{gathered}$

Number of combinations:

$\begin{gathered} C^8_6=\frac{8!}{(8-6)!6!} \\ =\frac{8!}{2!\times6!}=\frac{8\times7\times6!}{6!\times2\times1} \\ 6!\text{ cancel out, thus we have} \\ \frac{8\times7}{2} \\ \Rightarrow C_6^8=28 \end{gathered}$

Hence, there are 28 combinations and 20160 permutations.

7 0

1 year ago

I don't know how to do this

vesna_86 [32]

He could have 30 dimes and 13 quartars

7 0

3 years ago

The slope for 5 and 2

Yuri [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$