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3 years ago

Given: AD = BC and AD || BC

Mathematics

1 answer:

bearhunter [10]3 years ago

4 0

Answer:

ABCD is a parallelogram.

Step-by-step explanation:

A parallelogram is a quadrilateral that has two parallel and equal pairs of opposite sides.

From the given diagram,

Given: AD = BC and AD || BC, then:

i. AB = DC (both pairs of opposite sides of a parallelogram are congruent)

ii. <ADC = < BCD and < DAB = < CBA

thus, AD || BC and AB || DC (both pairs of opposite sides of a parallelogram are parallel)

iii. < BAC = < DCA (alternate angle property)

iv. Join BD, line AC and BC are the diagonals of the quadrilateral which bisect each other. The two diagonals are at a right angle to each other.

v. <ADC + < BCD + < DAB + < CBA = $360^{0}$ (sum of angles in a quadrilateral equals 4 right angles)

Therefore, ABCD is a parallelogram.

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What is the relationship between the slopes of two parallel lines?

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Answer:

Step-by-step explanation:

Parallel line slopes are always the same.

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3 years ago

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Prove that if one solution for a quadratic equation of theform x2 + bx + c = 0 is rational (where b and c are rational),then the

Elenna [48]

Answer:

The other solution of the given equation x² + bx + c = 0 is also rational number.

Step-by-step explanation:

Here, given: ax² + bx + c = 0 is a quadratic equation

Also, one solution (r) of equation is RATIONAL.

To show: The other solution (s) is also RATIONAL

Now, here: as x² + bx + c = 0

Since r and s are the two given solutions, the given equation can be factorized as:

x² + bx + c = (x -r) (x - s)

Simplifying LHS, we get:

(x -r) (x - s) = x x - r (x) - s (x) + (r)(s)

= x² + x(-r - s) + rs

or, x² + bx + c = x² + x(-r - s) + rs

Comparing the related terms, we get:

b = (-r - s)

⇒ b + s = - r

or, s = -r - b

Now, given : r = Rational and the negative of a rational is also rational.

⇒ -r is also rational

Also, difference of two rational number is also rational.

⇒ -r - b is also rational

⇒ s is a RATIONAL NUMBER

Hence, the other solution of the given equation x² + bx + c = 0 is also rational number.

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3 years ago

Jovani wanted to switch gyms. Fitness 19 charges a one time $39 cleaning fee and $17 per

Alexxx [7]

Answer:

Hi there!

Your answer is:

It will take 2.2 months for both gyms to be the same amount of money!

Step-by-step explanation:

Fitness 19:

39+ 17x

Anytime Fitness:

50+ 12x

Put them equal to each other:

39+17x = 50+12x

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39+5x = 50

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5x = 11

x= 2.2 months

39+17(2.2)= 76.4$

50+12(2.2)= 76.4$

Hope this helps!

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3 years ago

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How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$