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3 years ago

Given: ABCD is a trapezoid, AB = CD, BK ⊥ AD, AK = 10, KD = 20 Find: BC AD

Mathematics

1 answer:

alexdok [17]3 years ago

4 0

Answer:

 BC = 10 and AD = 30

Step-by-step explanation:

In figure-1 , AB = CD ,BK ⊥ AD, AK = 10, KD = 20.

Since, line AD is sum of AK and KD, then

AD = AK + KD

AD = 10 + 20

AD = 30

Since, BC ║AD and BK ⊥ AD then similarly we construct CL ⊥ AD

so, BC = KL and AK = LD

KL = AD - LD

KL = 20 - 10

KL = 10

Since, BC = KL then BC = 10

Hence,  BC = 10 and AD = 30

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6.2megagrams/hectoliter to kilograms/deciliter

lesya [120]

Answer:

6.2 kilograms/ deciliters

Step-by-step explanation:

megagrams to kilograms

1 megagram = 1000 kilograms

hectoliter to deciliter

1 hectoliter = 1000 deciliter

Using conversion factors

6.2megagrams 1000 kilgrams 1 hectoliter

---------------------- * ----------------------- * -------------------------

hectoliter 1 megagram 1000 deciliters

Canceling like terms

6.2 kilograms/ deciliters

6 0

3 years ago

Read 2 more answers

What is 2.99 as a fraction

Serga [27]

2.99

2 is a whole number

0.99 = 99/100

your fraction is:

mixed number = 2 99/100
improper fraction = 299/100 (multiply whole number to denominator & then add numerator)

hope this helps

8 0

3 years ago

Read 2 more answers

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

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3 years ago

What is 0.18 repeating simplified

Radda [10]

As a fraction it’s, 2/11

8 0

3 years ago

A company negotiator claims that only 35% of union members will support a strike, but a union rep believes that the true percent

Softa [21]

Answer:

$(d)$ $H_o: p = 0.35$ $H_a: p > 0.35$

Step-by-step explanation:

Given

$\mu = 35\%$

Required

The hypothesis statements

First claim: 35% will support a strike.

This represents the null hypothesis

$H_o: p = 0.35$

Second claim: A greater percentage will support the strike.

This represents the alternate hypothesis

$H_a: p > 0.35$

3 0

3 years ago