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2 years ago

These 3 please!!! Properties of Kites and Trapzoids exercises 1-3

Mathematics

1 answer:

scoundrel [369]2 years ago

7 0

Answer:

area = 1/2(AC)(BD)
yes; BE can be found using the Pythagorean theorem
yes

Step-by-step explanation:

Triangles ADB and CDB are congruent, so each is half the total area. The area of ADB is ...

A = 1/2bh

A = 1/2(DB)(1/2AC) = 1/4(DB)(AC)

The area of the kite is twice this value:

A = 1/2(DB)(AC)

In order to use the formula derived in part 1, we need to know the length of BE. That can be found using the Pythagorean theorem:

BE² +AE² = AB²

We know that AE = 1/2AC, so the desired dimension is ...

BE = √(AB² -(1/4)AC²)

Then the area of the kite is ...

area = 1/2(DB)(AC)

area = (1/2)(AC)(ED +√(AB² -(1/4)AC²)

Given three sides of a triangle, the area of it can be found using Heron's formula. If you have not yet seen that in your geometry class, you can always use the Pythagorean theorem and some algebra.

As before, we need to know the lengths of the diagonals. One is given. The other can be found considering the Pythagorean relations:

AD² = AE² +DE²

AB² = AE² +BE²

BE +DE = BD . . . . the sum of the horizontal diagonal lengths

Subtracting the second equation from the first gives ...

AD² -AB² = DE² -BE²

Using the third to write an expression for DE, we get an equation for BE:

AD² -AB² = (BD -BE)² -BE²

AD² -AB² = BD² -2BD·BE +BE² -BE²

(AB² +BD² -AD²)/(2BD) = BE

Substituting this into the original second equation, we can find AE, hence the area of the kite.

AE² = AB² -((AB² +BD² -AD²)/(2BD))²

Which means the kite area is ...

area = BD·AE = BD√(AB² -((AB² +BD² -AD²)/(2BD))²)

area = (1/2)√(4·AB²·BD² -(AB² +BD² -AD²)²

This area formula demonstrates it is possible to find the area of the kite given the three lengths AB, BD, DA.

_____

<em>Additional comment</em>

We have no doubt that the area formula of Part 3 can be rearranged to match Heron's formula. Here, we're interested in demonstrating area can be found.

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You own three different rings. You wear all three rings, but no two of the rings are on the same finger, nor are any of them on

Eddi Din [679]

Answer:

Therefore you can wear your rings in =336 ways

Step-by-step explanation:

Multiplication Law: If one occurs in x ways and second event occurs in y ways.Then the number of ways that two event occur in sequence is xy

Given that you have 3 different rings.

We have total 10 figures.

But you don't wear ring on your thumbs.

We have 2 thumbs.

So you can wear rings on (10-2) = 8 figures.

The ways of wearing of first ring is = 8

The ways of wearing of second ring is = 7

The ways of wearing of third ring is = 6

Therefore you can wear your rings in =(8×7×6)

=336 ways

4 0

3 years ago

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

5 0

2 years ago

65% convert to a fraction

I am Lyosha [343]

Answer:

$\frac{13}{20}$

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

How would you solve these 2 maths problems?

sweet-ann [11.9K]

Answer:

see explanation

Step-by-step explanation:

(9)

Expand and simplify right side and compare coefficients of like terms on left side.

(x - 2)(x² + ax + b) + c

= x³ + ax² + bx - 2x² - 2ax - 2b + c

= x³ + x²(a - 2) + x(b - 2a) - 2b + c

compare with

x³ + 2x² - 3x + 4

x² terms

a - 2 = 2 ( add 2 to both sides )

a = 4

x terms

b - 2a = - 3 ← substitute a = 4

b - 8 = - 3 ( add 8 to both sides )

b = 5

constant terms

- 2b + c = 4 ← substitute b = 5

- 10 + c = 4 ( add 10 to both sides )

c = 14

Thus a = 4, b = 5 and c = 14

-------------------------------------------------

(10)

Since both functions cross the x- axis at - 2 then (- 2, 0) satisfies both, that is

f(- 2) = $(-2)^{4}$ + a(- 2)³ + b(- 2)² + 36(- 2) + 144 = 0, that is

- 16 - 8a + 4b - 72 + 144 = 0

- 8a + 4b + 56 = 0

- 8a + 4b = - 56 → (1)

and

g(- 2) = $(-2)^{4}$ + (a + 3)(- 2)³ - 23(- 2)² + (b + 10)(- 2) + 40 = 0, that is

- 16 - 8(a + 3) - 92 - 2(b + 10) + 40 = 0

- 16 - 8a - 24 - 92 - 2b - 20 + 40 = 0

- 8a - 2b - 112 = 0

- 8a - 2b = 112 → (2)

Subtract (1) from (2) term by term

- 6b = 168 ( divide both sides by - 6 )

b = - 28

Substitute b = - 28 into (2)

- 8a + 56 = 112 ( subtract 56 from both sides )

- 8a = 56 ( divide both sides by - 8 )

a = - 7

Thus a = - 7 and b = - 28

5 0

3 years ago

What is the length of the hypotenuse, x, if (20,21,x) is a Pythagorean triple?

Sophie [7]

Answer:

Step-by-step explanation:

$(20,21,x) \\Pythagoras -Theorem =\\a^2+b^2 = c^2\\\\20^2 + 21^2 = x^2\\400 + 441 = x^2\\841 = x^2\\Square -root-both-sides\\\sqrt{841} = \sqrt{x^2} \\29 = x\\\\x = 29$

6 0

3 years ago

Read 2 more answers