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igor_vitrenko [27]

2 years ago

8

Find the value of x that makes the shape a rhombus.

1 answer:

goldenfox [79]2 years ago

5 0

Given:

CDEF is a rhombus.

To find:

The value of x.

Solution:

We know that the diagonals of a rhombus are angle bisectors because they bisect each vertex angle.

$5x+1=8x-20$

$5x-8x=-1-20$

$-3x=-21$

Divide both sides by -3.

$x=\dfrac{-21}{-3}$

$x=7$

Therefore, the value of x is 7.

You might be interested in

The cost of providing water bottles at a high school football game is $35 for the rental of the coolers and $0.50 per bottle

zloy xaker [14]

The cost function is an illustration of a linear function

The equation of the cost function is C(x) = 35 + 0.5x

The given parameters are:

$Rentals = 35$ -- the cost of renting a cooler

$Rate = 0.5$ --- the rate per bottle

Assume the number of bottles in a cooler is x.

The cost function for the bottles in a cooler would be:

$C(x) = Rentals + Rate \times x$

So, we have:

$C(x) = 35 + 0.5\times x$

Evaluate the products

$C(x) = 35 + 0.5x$

Hence, the equation is $C(x) = 35 + 0.5x$

Read more about linear functions at:

brainly.com/question/15602982

7 0

2 years ago

A pair of fair dice is cast. Let Edenote the event that the number landing uppermost on the first die is a 1 and let Fdenote the

iris [78.8K]

Answer:

They are not independent

Step-by-step explanation:

Given

E = Occurrence of 1 on first die

F = Sum of the uppermost occurrence in both die is 5

Required

Are E and F independent

First, we need to list the sample space of a roll of a die

$Event\ 1 = \{1,2,3,4,5,6\}$

Next, we list out the sample space of F

$Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\$

$6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}$

In (1): the sample space of E is:

$E = \{1\}$

So:

$P(E) = \frac{n(E)}{n(Event\ 1)}$

$P(E) = \frac{1}{6}$

In (2): the sample space of F is:

$F = \{5,5,5,5\}$

So:

$P(F) = \frac{n(F)}{n(Event\ 2)}$

$P(F) =\frac{4}{36}$

$P(F) =\frac{1}{9}$

For E and F to be independent:

$P(E\ and\ F) = P(E) * P(F)$

Substitute values for P(E) and P(F)

This gives:

$P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}$

$P(E\ and\ F) = \frac{1}{54}$

However, the actual value of P(E and F) is 0.

This is so because $E = \{1\}$ and $F = \{5,5,5,5\}$ have 0 common elements:

So:

$P(E\ and\ F) = 0$

Compare $P(E\ and\ F) = \frac{1}{54}$ and $P(E\ and\ F) = 0$ .

These values are not equal.

Hence: the two events are not independent

6 0

2 years ago

Which are the ordered pairs?

Aleonysh [2.5K]

I would say the answer is A and E

3 0

3 years ago

Solve the following equation for n 18n^2-50=0

Firdavs [7]

$18n^2-50=0\ \ \ |+50\\\\18n^2=50\ \ \ |:18\\\\n^2=\dfrac{50}{18}\\\\n^2=\dfrac{50:2}{18:2}\\\\n^2=\dfrac{25}{9}\to n=\pm\sqrt{\dfrac{25}{9}}\\\\n=\pm\dfrac{\sqrt{25}}{\sqrt{9}}\\\\\boxed{n=-\dfrac{5}{3}\ \vee\ n=\dfrac{5}{3}}$

$Used:\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$

3 0

3 years ago

Yall shiuld like kinda help...<br> youll be marked as “brainliest”

serg [7]

I’m getting (8 3/10 )
The picture explains more into depth

3 0

3 years ago