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3 years ago

The diagonals of rhombus abcd intersect at point

2 answers:

3 0

The area of a rhombus is 1/2 times the lengths of the diagonals

A = (1/2)(AC)(BD)

Since the diagonals bisect each other, and ed = 8, BD must be 16. Substituting in this for BD and the given area (168) we get the following:

168 = (1/2)(AC)16
168 = 8(AC)
21 = AC

Reptile [31]3 years ago

3 0

Answer:

ac= 21

Step-by-step explanation:

Rhombus is a parallelogram in which diagonals bisect each other .

Given: Area of rhombus is 168 .

The diagonals of rhombus abcd intersect at point e. It's diagonals are ac and bd . We know that diagonals of rhombus bisect each other ,so, bd=2de . So, $bd=2(8)=16$ .

Also, area of rhombus is $A=\frac{1}{2}d_1d_2$ where $d_1,d_2$ are the diagonals of rhombus .

As area of rhombus is 168 , we get ,

$168=\frac{1}{2}d_1d_2$

Let $d_1=bd=16$

So, we get ,

$168=\frac{1}{2}d_1d_2\\168=\frac{1}{2}\left ( 16 \right )d_2\\168=8d_2\\d_2=\frac{168}{8}\\=21$

Therefore, $ac=21$

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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

3 years ago

Find the least common denominator for these two rational expressions. B/b^2 -64 -7b/b^2+7b-8

san4es73 [151]

Answer:

The least common denominator is (b-8)(b+8)(b-1)

Step-by-step explanation:

We are given expression as

$\frac{b}{b^2-64}-\frac{7b}{b^2+7b-8}$

Firstly, we will factor both denominators

$b^2-64=b^2-8^2=(b-8)(b+8)$

$b^2+7b-8=(b+8)(b-1)$

so, we can plug it back

$\frac{b}{(b-8)(b+8)}-\frac{7b}{(b+8)(b-1)}$

First term denominator is

(b-8)(b+8)

Second term denominator is

(b+8)(b-1)

So,

Least common denominator will be

(b-8)(b+8)(b-1)

So, we get

$LCD=(x-8)(x+8)(x-1)$

3 0

2 years ago

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

The sum of two numbers is 62 the smaller number is 6 less than the larer number what are the numbers

muminat

X= larger number
x-6= smaller number

x + (x - 6)= 62
combine like terms

2x - 6= 62
add 6 to both sides

2x= 68
divide both sides by 2

x= 34 larger number

SMALLER NUMBER
x - 6= 34 - 6= 28

ANSWER: The smaller number is 28 and the larger number is 34.

Hope this helps! :)

5 0

3 years ago

A pizza has a diameter of 12 inches. Each piece has a central angle of radians. What is

Firlakuza [10]

Answer:

The area of one slice of pizza is 18 in²

Step-by-step explanation:

First we have to know that the 360 ° representing the circle expressed in radians are expressed as 2π

Now we must calculate the area of the circle

a = π * r²

a = π * (12in)²

a = 144π in²

Now that we have the area of the circumference we multiply it by (π/4) / (2π)

144π in² * (π/4) / (2π) =

144 in² * 1/8 =

18 in²

The area of one slice of pizza is 18 in²

3 0

3 years ago