All categories

3 years ago

What is the value of s and the length of side BC if ABCD is a rhombus?

Mathematics

2 answers:

OverLord2011 [107]3 years ago

5 0

The rhombus has four equal sides

10s - 16 = 9s + 29
10s - 9s = 29 + 16
s = 45

BC = CD = 10s-16 = 10*45 - 16 = 434

Natali5045456 [20]3 years ago

4 0

Step 1

Find the value of s

we know that

In a Rhombus all sides are congruent

$AB=BC=CD=DA$

$AB=9s+29\\CD=10s-16$

equate AB and CD

$9s+29=10s-16\\$

Combine like term

$10s-9s=29+16\\$

$s=45\ units$

The answer Part a) is

the value of s is $45\ units$

Step 2

Find the value of side AB

$AB=9s+29$

substitute the value of s

$AB=9*45+29=434\ units$

Remember that the sides are congruent

$AB=BC=CD=DA$

therefore

the answer Part b) is

The length of the side BC is $434\ units$

You might be interested in

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago

Factor completely x^2-9x-20

Nadya [2.5K]

Answer:

(x - 10.844) (x + 1.844)

Step-by-step explanation:

5 0

3 years ago

PLEASE HELP. Given the polynomial 6x4+ 3x3- 7x2 + 4x + 5, choose ALL the following statements that are true?

Lunna [17]

Answer:

The polynomial has a leading coefficient of 4

The polynomial has 4 terms

The polynomial has a degree of 4

those are the answers

Step-by-step explanation:

its just there

6 0

3 years ago

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y

Anestetic [448]

Answer:

$\frac{1}{4180}$

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are $12$ red candies, $12$ blue, $7$ green, $13$ brown, $3$ orange, and $10$ yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total candies in the bag $=57$

Probability that first ball is orange, $P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{3}{57}=\frac{1}{19}$

Probability that second ball is orange, $P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{2}{56}=\frac{1}{28}$

Probability that third ball is green, $P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}$

$=\frac{7}{55}$

Now, probability that first two balls are orange and third is green is

$=P(A)\times P(B)\times P(C)$

$=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}$

$=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}$

$=\frac{1}{4180}$

Hence, probability that first two balls are orange and third is green is $\frac{1}{4180}$

3 0

3 years ago

While testing a new vaccine, a lab technician noticed that the number of bacteria reduced by half every minute. How long does it

Kaylis [27]

It takes three minutes for the bacteria to get to 375. to do this you put the equation 3000÷2^x, (^x means to the X power) where X is the amount of minutes, and obviously 3000 is the start amount.

4 0

3 years ago