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3 years ago

Given ABCD is a parallelagram diagonals AC and BD intersect at E Prove AE is congruent to CE and BE is congruent to DE

Mathematics

2 answers:

Zepler [3.9K]3 years ago

8 0

In the parallelogram ABCD, diagonals AC and BD intersect at point E.

According to the definition of parallelogram, opposite sides are equal and parallel to each other. That means, AB = DC

Now as AB and DC are parallel, so according to the property of Alternate Interior Angles, we will get:

∠EAB = ∠ECD and ∠EBA = ∠EDC

Thus , in two triangles ΔABE and ΔDCE, two angles and one side are equal. So, ΔABE and ΔDCE are congruent to each other.

That means, AE = CE and BE = DE

So, AE is congruent to CE and BE is congruent to DE

Sergio039 [100]3 years ago

8 0

1. ABCD is a parallelogram --Given

2. AB≌CD--parallelogram side theorem

3. AB∥CD--def. of parellelogram

4. ∠ABE and ∠CDE are alt. interior angles-- def. of alt. interior angles

5.∠BAE and ∠DCE are alt. interior angles-- def. of alt. interior angles

6. ∠BAE≌∠DCE--alt. interior angles theorem

7. ∠ABE≌CDE--alt. interior angle theorm

8. ⊿BAE≌⊿DCE-- ASA

9. AE≌CE-- CPCTC

10. BE≌DE-- CPCTC

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3 years ago

1. What would the polynomial x + 7 – 4x^3 be written as in standard form?

yarga [219]

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3 years ago

Exterior of scalene triangle sides are 25 and 15 and x. solve x.

Lyrx [107]

-- All three sides of a scalene triangle have different lengths.
So 'x' can't be 15 and it can't be 25.

-- 'x' must be 10 or more in order to reach between the ends
of the 25 and the 15.

-- 'x' must be less than 40 in order for the 25 and the 15 to reach
between its ends.

So the value of 'x' must satisfy these conditions:

0 < x < 15
15 < x < 25
25 < x < 40

Any number that satisfies these conditions is an acceptable value for 'x'.

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3 years ago

Read 2 more answers

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Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

3 years ago

Miguel's cat, Chester, weighs 6.8 pounds. Chester's vet tells Miguel to feed the cat 3 4 of an ounce of canned food daily for ea

Vitek1552 [10]

Answer:

0.4 ounces

Step-by-step explanation:

In a day

Miguel is to feed Chester ,

Chester can eat 3/4 of an ounce for each pound of body weight.

Chester weighs 6.8 pounds

Hence:

1 pound = 3/4 ounce

6.8 pounds = x ounces

Cross Multiply

x = 6.8 × (3/4) ounces

x = 5.1 ounces

Chester is to be fed 5.1 ounces of food each day.

The amount of food left after 1 day

= 5.5 ounces - 5.1 ounces

= 0.4 ounces

Hence, 0.4 ounces of food is left after 1 day

7 0

2 years ago